Hi everyone,
I am going to make 0.1 M of sodium lactate for Manganese peroxide assay at pH 4.5. Does any one can help me in the calculation of each components?
Many thanks in advance.
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I hope that you will benefit in the Moularia account
https://www.learnchemistry12.com/2017/01/M-concentration.html
pH =pKa + log [sod. lactate]/[lactic acid];
4.5-3.86= log [sod. lactate]/[lactic acid]= 0.64
The pH of a sodium lactate aq. sol. can be predicted as follows:
The pH of the NaC3H5O3 (sodium lactate) aq. sol. is expected to be weakly basic (*) and chiefly dictated by hydrolysis of the salt anion: (C3H5O3)- + H2O ⇌ C3H6O3 + OH-, Kh = [C3H6O3]·[OH-]/[(C3H5O3)-] = Kw/Ka, where we may take Ka = [(C3H5O3)-]·[H3O+]/[C3H6O3] = 1.38·10-4 M (pKa = 3.86), while Kw = 1.0·10-14 M2 is the autodissociation constant of water. Hence: [C3H6O3] = [(C3H5O3)-]·[H3O+]/Ka.
Sodium molar balance writes: CNaC3H5O3 = [Na+] + [NaC3H5O3], where CNaC3H5O3 is the nominal molar concentration of dissolved salt (formality, to be precise). Sodium lactate can reasonably be considered highly dissociated in aq. sol.: [NaC3H5O3] 5.86) and [H3O+] 100·Ka = 1.38·10-2 M, we may write: pH ≈ 8.93 + (½)log10(CNaC3H5O3/M). Consequently, for a 0.1 M sodium lactate sol., we can predict pH ≈ 8.4.
(*) Is the salt of a (relatively) weak acid and a strong base.
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