Question #170828

A 0.10 M solution of formic acid (HCOOH) is prepared in various solvents: (a) pure water, (b) 0.10 M NaHCOO solution, (c) 0.10 M Ca(HCOO)2 solution, and (d) 0.10 M NaCl. Compute the pH of each solution. The Ka of formic acid is 1.78*10^-4.


1
Expert's answer
2021-03-12T06:29:26-0500

a)pH=1/2(pKalog(c0))=2.38pH=1/2(pK_a-log(c_0))=2.38

b)pH=pKa+log(c(NaHCOO)c(HCOOH))=pKa=3.75pH=pK_a+log\left(\frac{c(NaHCOO)}{c(HCOOH)}\right)=pK_a=3.75

c) pH=pKa+log(2c(Ca(HCOO)2)c(HCOOH))=pKa+log(2)=4.05pH=pK_a+log\left(\frac{2c(Ca(HCOO)_2)}{c(HCOOH)}\right)=pK_a+log(2)=4.05

d)pH=2.38pH=2.38


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