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 x 10
a)pH=1/2(pKa−log(c0))=2.38pHa)pH=1/2(pK_a-log(c_0))=2.38pHa)pH=1/2(pKa−log(c0))=2.38pH
b)pH=pKa+log(c(NaHCOO)c(HCOOH))=pKa=3.75pHb)pH=pK_a+log\left(\frac{c(NaHCOO)}{c(HCOOH)}\right)=pK_a=3.75pHb)pH=pKa+log(c(HCOOH)c(NaHCOO))=pKa=3.75pH
c)pH=pKa+log(2c(Ca(HCOO)2)c(HCOOH))=pKa+log(2)=4.05pHc) pH=pK_a+log\left(\frac{2c(Ca(HCOO)_2)}{c(HCOOH)}\right)=pK_a+log(2)=4.05pHc)pH=pKa+log(c(HCOOH)2c(Ca(HCOO)2))=pKa+log(2)=4.05pH
d)pH=2.38d) pH =2.38d)pH=2.38
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