The pH of water is generally basic ie. 7. Therefore diluting by 20X to reach 50mM would decrease the pH of your buffer. This is not uncommon. Just re-pH after you've diluted. If possible, try adding HEPES 20mM to your solution to keep the pH stable. 

Dear Patryk,

you can simply calculate the effect using the Henderson-Hasselbalch equation for buffers with one pKa value and using the law of mass action in general, also for polyprotic systems.

But you even do not need formulas to explain the reason for your observation. Adding water to your acidic system means two things, 1st you are diluting your buffer, thus weakening its buffering capacity and 2nd you are adding a comparably basic liquid (as Kimberly already mentioned, water usually is at pH 7 or higher). So it's quite obvious that the pH of your solution will increase slightly.

Keep in mind that pH is temperature dependent, this might also contribute to your observation. Citric acid has a considerablle hydration enthalpy. So dissolving it or its salts will change the temperature of your stock solution compared to the temperature after dilution, and thus influence the corresponding pH measurements. The effect is even stronger if you use a stock solution coming from a 4C fridge and dilute it with water at room temperature.

Hope that helps...

The CRC lists the published values of 1:2 dilutions of many buffers.  The values do change and the diluted values are well known in many cases.

About predicting the pH of citric acid ― sodium citrate solutions and buffers for approx. 4.0 < pH < 5.5:

Citric acid (H3C6H5O7 ≡ H3Cit) is a tribasic weak acid. Sodium citrate salts (Na3Cit; Na2HCit; NaH2Cit) can reasonably be considered highly dissociated in aq. sol. (strong electrolytes). It can be noticed ― based on the speciation diagram for citric acid ― that for the pH range of interest, say 4.0 < pH < 5.5, we can neglect the contribution of species Cit3- and H3Cit. For pH prediction, is then enough to consider just the 2nd acid dissociation equilibrium and the related constant:

H2Cit- + H2O ⇌ HCit2- + H3O+; Ka2 = [HCit2-][H3O+]/[H2Cit-] = 1.74·10-5 M; pKa2 = 4.76; [H2Cit-] ≠ 0 M. Hence: [H2Cit-] = [HCit2-][H3O+]/Ka2.

Citrate molar balance then writes: CCit ≈ [HCit2-] + [H2Cit-] ≠ 0 M

Here CCit is the joint nominal (formal, to be precise) molar concentration of dissolved citrate; either as citric acid or as citrate salt(s) ― C generically denotes formal concentrations (M units): CCit ≡ CH3Cit + CNa3Cit + CNa2HCit + CNaH2Cit ≠ 0 M.

Hence:

CCit ≈ [HCit2-] + [HCit2-][H3O+]/Ka2

So (*):

[HCit2-] ≈ CCit /{1 + [H3O+]/Ka2}; [H2Cit-] ≈ CCit[H3O+]/{Ka2 + [H3O+]}

Sodium molar balance writes:

CNa = [Na+] (*); where CNa is the joint nominal molar concentration of dissolved sodium as any citrate salt, and possibly, also as NaOH: CNa ≡ 3CNa3Cit + 2CNa2HCit + CNaH2Cit + CNaOH.

Charge balance writes: [Na+] + [H3O+] ≈ [OH-] + 2[HCit2-] + [H2Cit-]

After substitution (*), this rewrites as:

CNa + [H3O+] ≈ Kw/[H3O+] + CCit{2/{1 + [H3O+]/Ka2} + [H3O+]/{Ka2 + [H3O+]}}

or (**):

[H3O+]2 + [H3O+]·{CNa - CCit{1 + Ka2/{Ka2 + [H3O+]}}} ≈ Kw

Example I:

For a mixed equimolar salt solution with Cs ≡ CNa2HCit = CNaH2Cit; with CH3Cit = CNaOH = CNa3Cit = 0 M; CCit ≡ CNa2HCit + CNaH2Cit; CNa ≡ 2CNa2HCit + CNaH2Cit; CCit ≡ CNa2HCit + CNaH2Cit = 2Cs; CNa ≡ 2CNa2HCit + CNaH2Cit = 3Cs

Then:

[H3O+]2 + [H3O+]·Cs{1 - 2Ka2/{Ka2 + [H3O+]}} ≈ Kw

or:

Cs{2Ka2/{Ka2 + [H3O+]} - 1} ≈ [H3O+] - [OH-]

where, for the expected acidic pH, we should have:

2Ka2/{Ka2 + [H3O+]} > 1; 2Ka2 > Ka2 + [H3O+]; [H3O+] < Ka2; log10{[H3O+]/M} < log10{Ka2/M}; pH > pKa2 = 4.76.

Approximate solution is possible from last equation:

2CsKa2 ≈ {Cs + [H3O+] - Kw/[H3O+]}·{Ka2 + [H3O+]}

for not too dilute acidic solutions, with Cs > 100[OH-] = 100Kw/[H3O+] and [H3O+] > 100[OH-]:

2CsKa2 ≈ {Cs + [H3O+]}·{Ka2 + [H3O+]}

[H3O+]2 + (Ka2 + Cs)·[H3O+] ≈ CsKa2

what solves as:

[H3O+] ≈ - ½(Ka2 + Cs) + √{CsKa2 + ¼(Ka2 + Cs)2}

From this we predict that pH ≈ 4.76 as long as Cs > 2.7·10-3 M, while pH noticeably increases just for 100Kw/[H3O+] < Cs < 2.7·10-3 M.

Example II:

For (pure) citric acid aq. sol. (CNa = 0 M; CCit = CH3Cit ≠ 0 M):

[H3O+] - CH3Cit·{1 + Ka2/{Ka2 + [H3O+]}} ≈ Kw/[H3O+]

because the solution is sufficiently acidic; so that we expect [H3O+] > 100Kw/[H3O+] = 100[OH-] and CH3Cit·{1 + Ka2/{Ka2 + [H3O+]}} > 100[OH-]:

[H3O+]2 - (CH3Cit - Ka2)·[H3O+] ≈ 2CH3Cit·Ka2

what solves as:

[H3O+] ≈ ½(CH3Cit - Ka2) + √{2CH3Cit·Ka2 + ¼(CH3Cit - Ka2)2}

for sufficiently concentrate sol., with CH3Cit > 100Ka2 = 1.74·10-3 M:

[H3O+] ≈ ½(CH3Cit)·{1 + √(1 + 8Ka2/CH3Cit)}

for sufficiently concentrate sol., with 8Ka2/CH3Cit < 1/100, i.e. CH3Cit > 1.39·10-2 M:

[H3O+] ≈ CH3Cit; pH = - log10{CH3Cit/M} ≡ pCH3Cit

Inverse solution:

Generally speaking, there is no single solution of (**) for CNa and CCit ≠ 0 M; for known/fixed [H3O+] =10-pH M. Solution can be single if some convenient restrictions are applied to the substances actually involved. Otherwise, there will be different compositions that give the same solution and/or buffer.

Inverse solution (pH → [H3O+] → CH3Cit) for (pure) citric acid aq. sol. (CNa = 0 M; CCit = CH3Cit ≠ 0 M; from (**):

CH3Cit ≈ {[H3O+] - Kw/[H3O+]}·{Ka2 + [H3O+]}/{2Ka2 + [H3O+]}

Is this a phenomenon of increased ‚ionic’ dissociation ir only of the neutral pH of water? Usually one uses a. dest or - better - a. bidest to mix up the crystalline buffer ‚ingredients’. I‘ve (pH-) measured a. bi- and also a.tridest (produced by /or from various sources) and found differing values, most times between pH 6.2 and 6.8... in one case pH was measured with incredible 5.9.

‚Take-home-message‘: if you prepare buffer solutions (even using published protocols!) either as stock solutions (with higher molarity) AND - most importantly - diluting the latter to ‚work solutions‘ ALWAYS appropriate checks by mesurenent for pH AND osmolarity are recommebded

Answer #[008]

With regard to my answer [007] above which was visible as 'multiple answers' of the same content for a short time:

Apologies: Hoping to be able to delete all those multiple answers created by dysfunctional or incomprehensible / inexplicable APP (when posting the answer via iPhone I was not notified correctly about a "successfully sent" answer...

WHM

EDIT:

i) fortunately, deletion of all the multiple identical answers in excess was possible via access to the RG-Thread using my home-computer.

ii) IT IS NOT POSSIBLE (at least for me!) to edit answers sent / posted via iPhone and the RG-APP when trying to do so in the / a thread accessed via home-PC: the attempt is followed by changing thread display to "blank field" which is totally infunctional.

iii) not knowing how to 'delete' the reply posted by re-access via iPhone: cannot find a function-icon within the iPhone display, telling me: DELETE answer...etc).

iv) therefore I find the RG-APP (at least for me useless and will not try again to post an answer via iPhone or 'recommend' an answer - too - since i found out that such a recommendation will not appear in the RG-thread properly.

So finally - to edit my previous answer [007] (above) in this thread with the correct and intended text:

Corrected Answer #[007] by Wolfgang H. Muss (17th Mar, 2021 11.50 a.m.

"Is this a phenomenon of increased ‚ionic’ dissociation or only of the neutral pH of water? *#*) see bottom

Usually one uses 'a. dest' or - better - 'a. bidest' to mix up the crystalline buffer ‚ingredients’.

I‘ve (pH-) measured 'a. bi-' and also 'a. tridest' (produced by /or from various sources) and found differing values, most times between pH 6.2 and 6.8... in one case pH was measured with incredible 5.9.

‚Take-home-message‘: if you prepare buffer solutions (e.g. for the first second or third time- not having still sufficient experience - even using published protocols!) either as stock solutions (with higher molarity) AND - most importantly - diluting the latter to ‚work solutions‘ : ALWAYS appropriate checks by measurement for pH AND osmolarity are recommended.

*#*) NB-Only as an example: (Na-Na-)Phosphate buffer stock solution 2 M correctly is measured by an osmometer with xxx mosmol.

Diluted 1:1 with water (= 1 M) - by correct measurement with an osmometer - will NOT create a buffer solution with xxx/2 mosmol but a value a bit higher than xxx/2... similarly if 1M PO4 buffer is diluted 1:1 (=0.5M) again, it will not be a value half that of 1M, but a bit higher than half the value....this must be accounted for by higher "dissociation of PO4-ions ' in more diluted molarities...

Carlos Araújo Queiroz , thank you for the really impressive chemical notion...am I right, if I (as a non-chemist) suppose that you anticipate 'ionic dissociation' (achieved by a certain / different grade of dilution) too to be responsible for pH-decrement (as we can see this chemical behaviour also to be present in other buffer solutions....complicated additionally with variability in pH due to temperature...?)? Thank you for -eventually - a short elucidaton... Best regards, WHM

@ Gayle Gleichauf , regarding (QUOTE):

'The CRC lists the published values of 1:2 dilutions of many buffers.  The values do change and the diluted values are well known in many cases.' (END QUOTE)

My question to you: would it be possible to add an URL or a pdf of the CRC-List, just for the convenience of the reader??

THANK YOU...

(Disclaimer:no affiliation to THERMO_FISHER_Scientific, but in former times satisfied customer and 'consumer' of their product offerings (Vienna-Austria Branch).

Wolfgang H. Muss ― Considering your post above addressed to myself:

After my first post, above:

(i) «Sodium citrate salts (Na3Cit; Na2HCit; NaH2Cit) can reasonably be considered highly dissociated in aq. sol. (strong electrolytes)» ― I have (approximately) taken [Na3Cit] = [Na2HCit] = [NaH2Cit] = 0 M. The salts formal concentratons, respectively CNa3Cit, CNa2HCit, CNaH2Cit; may be not null. The same applies for [NaOH] = 0 M and CNaOH (may be not null); because NaOH is a strong base. Formal concentratons are usually restricted to neutral substances and do not account for speciation.

(ii) «for the pH range of interest, say 4.0 < pH < 5.5, we can neglect the contribution of species Cit3- and H3Cit» ― I have (approximately) taken [Cit3-] = [H3Cit] = 0 M. The corresponding salt and acid formal concentratons, respectively, CNa3Cit and CH3Cit, may be not null.

(iii) The 2nd acid dissociation equilibrium does not include any undissociated (neutral) substance.

So; the degree of 'ionic dissociation' was not considered at my post above. ― It does not influence reasoning. It would play a role if the 1st acid dissociation equilibrium was accounted for, as might be found justifiable for about pH < 4.0 ― according to the speciation diagram for citric acid (*).

About the temperature effect:

(iv) The acid dissociation constants are currently typically listed for 25 ºC. Both formal concentrations (because volumetric) and acid dissociation constants are temperature dependent.

(v) On the effect of temperature on acid-base equilibria of aqueous solutions ─ you may find useful to check: Herbert A. Laitinen, Walter E. Harris, “Chemical Analysis ─ An Advanced Text and Reference”, 2nd ed., McGraw-Hill, 1975, pp. 47-48.

―

(*) https://en.wikipedia.org/wiki/Citric_acid

Carlos Araújo Queiroz : Thank you so much !

Have a good and successful week,

best regards, WHM.

About predicting the pH of citric acid ― sodium citrate solutions and buffers for approx. pH < 4.0 ― cf. my post at:

https://www.researchgate.net/post/How-to-prepare-a-citric-acid-solution-of-pH-25-and-3-from-100-citric-acid-powder

About predicting the pH of citric acid ― sodium citrate solutions and buffers for approx. 5.5 < pH < 8.0:

Citric acid (H3C6H5O7 ≡ H3Cit) is a tribasic weak acid. Sodium citrate salts (Na3Cit; Na2HCit; NaH2Cit) can reasonably be considered highly dissociated in aq. sol. (strong electrolytes). It can be noticed ― based on the speciation diagram for citric acid ― that for the pH range of interest, say 5.5 < pH < 8.0, we can neglect the contribution of species H3Cit and H2Cit-. For pH prediction, is then enough to consider just the 3rd acid dissociation equilibrium and the related constant:

HCit2- + H2O ⇌ Cit3- + H3O+; Ka3 = [Cit3-][H3O+]/[HCit2-] = 4.0·10-7 M; pKa3 = 6.40; [HCit2-] ≠ 0 M. Hence: [HCit2-] = [Cit3-][H3O+]/Ka3.

Citrate molar balance then writes: CCit ≈ [HCit2-] + [Cit3-] ≠ 0 M.

Here, CCit is the joint nominal (formal, to be precise) molar concentration of dissolved citrate; either as citric acid or as citrate salt(s) ― C generically denotes formal concentrations (M units): CCit ≡ CH3Cit + CNa3Cit + CNa2HCit + CNaH2Cit ≠ 0 M. Hence:

CCit ≈ {[H3O+]/Ka3 + 1}·[Cit3-]

So (*): [Cit3-] ≈ CCit/{[H3O+]/Ka3 + 1}; [HCit2-] = (CCit/Ka3)[H3O+]/{[H3O+]/Ka3 + 1}

Sodium molar balance writes: CNa = [Na+] (*); where CNa is the joint nominal molar concentration of dissolved sodium as any citrate salt, and possibly, also as NaOH: CNa ≡ 3CNa3Cit + 2CNa2HCit + CNaH2Cit + CNaOH.

Charge balance writes: [Na+] + [H3O+] ≈ [OH-] + 2[HCit2-] + 3[Cit3-]

After substitution (*), this rewrites as (**):

[H3O+]3 + {Ka3 + (CNa - 2CCit)}·[H3O+]2 + ((CNa - 3CCit)Ka3 - Kw)·[H3O+] ≈ KwKa3

or, with ΔC12 ≡ CNa - 2CCit; ΔC13 ≡ CNa - 3CCit:

[H3O+]3 + (Ka3 + ΔC12)·[H3O+]2 + (ΔC13Ka3 - Kw)·[H3O+] ≈ KwKa3

Example I:

For equimolar sol. of Na2HCit and Na3Cit sol., thus with CH3Cit = CNaH2Cit = CNaOH = 0; Cs ≡ CNa2HCit = CNa3Cit; CCit ≡ 2Cs; CNa ≡ 5Cs; with ΔC12 ≡ 5Cs - 4Cs = Cs; ΔC13 ≡ 5Cs - 6Cs = - Cs:

[H3O+]3 + (Ka3 + Cs)·[H3O+]2 - (CsKa3 + Kw)·[H3O+] ≈ KwKa3

For not too dilute sol.; CsKa3 > 100Kw and Cs > 100Ka3:

Cs{[H3O+] - Ka3}[H3O+] ≈ KwKa3 - [H3O+]3

For not too dilute sol.; Cs{[H3O+] - Ka3}[H3O+] > 100{KwKa3 - [H3O+]3}:

[H3O+] ≈ Ka3; pH ≈ pKa3

Refined solution (iterative):

{[H3O+]i+1}2 - 2½Ka3[H3O+]i+1 ≈ {KwKa3 - [H3O+]i3}/Cs

what solves as:

[H3O+]i+1 ≈ ½Ka3 + ½Ka3√{1 + 4{Kw/Ka3 - [H3O+]i3/(Ka3)2}/Cs}

where we can take:

[H3O+]0 = Ka3

with (e.g.) Cs = 1.0·10-1 M; [H3O+]1 ≈ ½Ka3{1+ √(1 - 3.0·10-6)} ≈ Ka3.

Example II:

For (pure) citric acid aq. sol. (CNa = 0 M; CCit = CH3Cit ≠ 0 M; ΔC12 ≡ - 2CH3Cit;

ΔC13 ≡ - 3CH3Cit; approx. 5.5 < pH < 7.0:

[H3O+]3 + (Ka3 - 2CH3Cit)·[H3O+]2 - (3CH3Cit·Ka3 + Kw)·[H3O+] ≈ KwKa3

For not too dilute sol.; with (approx. i) 3CH3Cit·Ka3 > 100Kw and (approx. ii) 2CH3Cit > 100Ka3:

[H3O+]3 - 2CH3Cit[H3O+]2 - (3CH3Cit·Ka3)·[H3O+] ≈ KwKa3 = [OH-]·[H3O+]Ka3

or:

[H3O+]2 - 2CH3Cit[H3O+] ≈ {[OH-] + 3CH3Cit}·Ka3

For not too dilute sol.; with (approx. iii) 3CH3Cit > 100[OH-]:

[H3O+]2 - 2CH3Cit[H3O+] ≈ 3CH3Cit·Ka3

what solves as:

[H3O+] ≈ CH3Cit·{1 + √(3Ka3/CH3Cit + 1)}

with (approx. iv) 3Ka3/CH3Cit < 1/100:

[H3O+] ≈ 2CH3Cit; pH = - log10{[H3O+]/M} ≈ pCH3Cit - 0.30, with pCH3Cit ≡ - log10(2CH3Cit/M).

Inverse solution:

Generally speaking, there is no single solution of eq. (**) for CNa and CCit ≠ 0; for known/fixed [H3O+] =10-pH M. Solution can be single if convenient restrictions are applied to the substances actually involved. Otherwise, there will be different compositions that give the same solution and/or buffer.

Inverse solution (pH → [H3O+] → CH3Cit) for (pure) citric acid aq. sol. (CNa = 0 M; CCit = CH3Cit ≠ 0 M; approx. 5.5 < pH < 7.0); from (**):

CH3Cit ≈ {[H3O+]2 - Kw}·{Ka3 + [H3O+]}/{[H3O+]·{Ka3 + 2[H3O+]}}

About predicting the pH of citric acid ― sodium citrate solutions and buffers for approx. pH > 8.0:

Much of the above post reasoning (refer to above post, dedicated to approx. 5.5 < pH < 8.0) still applies for higher pH. However, for this basic pH range, [OH-] > 100[H3O+]; and we can simplify charge balance, as:

[Na+] ≈ [OH-] + 2[HCit2-] + 3[Cit3-] > 0 M

After substitution (as in above post):

CNa ≈ Kw/[H3O+] + 2(CCit/Ka3)·[H3O+]/{[H3O+]/Ka3 + 1} + 3CCit/{[H3O+]/Ka3 + 1} > 0 M

or (**):

{(CNa - 2CCit)/Ka3}·[H3O+]2 + (CNa - 3CCit - Kw/Ka3)·[H3O+] ≈ Kw

or with ΔC12 ≡ CNa - 2CCit ≠ 0; ΔC13 ≡ CNa - 3CCit:

[H3O+]2 + (Ka3/ΔC12)·(ΔC13 - Kw/Ka3)·[H3O+] ≈ Ka3Kw/ΔC12

what solves as:

[H3O+] ≈ - ½(Ka3/ΔC12)·(ΔC13 - Kw/Ka3) + √{Ka3Kw/ΔC12 + ¼(Ka3/ΔC12)2·(ΔC13 - Kw/Ka3)2}

with ΔC12 > 0 M, since [H3O+] > 0 M.

Example I:

For pure Na3Cit sol. with CCit = CNa3Cit = 1.0·10-2 M; CNa = 3CNa3Cit; ΔC12 = 3CNa3Cit - 2CNa3Cit = CNa3Cit; ΔC13 = 3CNa3Cit - 3CNa3Cit = 0 M. Hence: [H3O+] ≈ ½(Kw/CNa3Cit){1 + √(1 + 4Ka3CNa3Cit/Kw)} = ½((1.0·10-14 M2)/(1.0·10-2 M))·{1 + √(1 + 4(4.0·10-7 M)·(1.0·10-2 M))/(1.0·10-14 M2))} = 6.35·10-10 M; pH = 9.20.

Example II:

For pure equimolar Na3Cit and NaOH sol. with C ≡ CNa3Cit = CNaOH = 1.0·10-2 M; CCit ≡ CNa3Cit = C; CNa ≡ 3CNa3Cit + CNaOH = 4C: ΔC12 ≡ CNa - 2CCit = 2C; ΔC13 ≡ CNa - 3CCit = C. Hence:

[H3O+] ≈ - ¼(Ka3/C)·(C - Kw/Ka3) + √{½Ka3Kw/C + ¼(½Ka3/C)2·(C - Kw/Ka3)2} ≈ ¼Ka3{ - 1+ √{1 + 8Kw/(C·Ka3)}} = ¼Ka3{ - 1+ √{1 + 2(1.0·10-7 M)/C}} = ¼Ka3{ - 1+ √{1 + 2·10-5}} = ¼(4.0·10-7 M)·10-5 = 1.0·10-12 M; pH = 12.00. Solution equals that of Example III (see below).

Inverse solution:

Generally speaking, there is no single solution of (**) for CNa > 0 M and CCit ≠ 0 M; for known/fixed [H3O+] =10-pH M. Solution can be single if some convenient restrictions are applied to the substances actually involved. Otherwise, there will be different compositions that give the same solution and/or buffer.

Simplified approximate solution for approx. pH > 9.0:

It can be obtained by neglecting the contribution of HCit2- and of the acid 3rd acid dissociation equilibrium (besides 1st and 2nd) ― as supported by the speciation diagram for citric acid. That can be considered approximately valid if moderate excess of NaOH remains (free) in solution, as long as the pH contribution from this strong base can be considered much higher compared with that of citrate (Cit3- ) hydrolysis.

It was shown (above post; *) that: [HCit2-] = (CCit/Ka3)[H3O+]/{[H3O+]/Ka3 + 1} = (CCit/Ka3)/{1/Ka3 + 1/[H3O+]}. For pH > 9; [H3O+] < 10-9 M; [HCit2-] < CCit/401.

It was also shown that: [Cit3-] ≈ CCit/{[H3O+]/Ka3 + 1}. For pH > 9; [Cit3-] > CCit/4.04.

Hence, for pH > 9; [Cit3-]/[HCit2-] > (CCit/4.04)/(CCit/401) = 401/4.04 = 99.3 ≈ 102.

We can hence consider to further simplify charge balance, as:

[Na+] ≈ [OH-] + 3[Cit3-]

Citrate molar balance now writes: CCit ≈ [Cit3-] ≠ 0 M.

Hence: CNa ≈ [OH-] + 3CCit; [OH-] = Kw/[H3O+] ≈ CNa - 3CCit ≠ 0 M.

[H3O+] ≈ Kw/(CNa - 3CCit); pH = - log10{[H3O+]/M} = pKw + log10{(CNa - 3CCit)/M}.

Example III:

For pure equimolar Na3Cit and NaOH sol. with C ≡ CNa3Cit = CNaOH = 1.0·10-2 M; CCit ≡ CNa3Cit = C; CNa ≡ 3CNa3Cit + CNaOH = 4C: [H3O+] ≈ Kw/C = ; pH = pKw + log10{1.0·10-2} = 12.00. Solution equals that of Example II.

To prepare the 0.1 ml Citrate Buffer PH 4.5, for solution dissolved 29.4mg/ml in distilled water. they need for 500microlitre of this buffer but made a larger volume, so it is easier to PH buffer of adjusted 4.5