Question

N/10 acetic acid was titrated with N/10 Naoh

Answer 1

Answer:

$n_{f}$ for both $CH_{3} COOH$ and NaOH=1

So normality=molarity

$\frac{N}{10} =\frac{1}{10} =0.1M Acetic acid$ and 0.1M NaOH

$CH_{3} COOH+NaOH---->CH_{3} COO^{-} Na^{+} +H_{2} O$

At t=0                0.1                   0.1

At t2=25%     0.1-0.025      0.1-0.025

                      0.075              0.075

at t3=50%     0.1-0.050      0.1-0.050

                      0.050              0.050

at t4=75%      0.1-0.075      0.1-0.075

                      0.025               0.025

pH=$-logK_{a} +log\frac{[salt]}{[Acid]}$

Here

$-logK_{a} =-log10^{-5}$

                      =+5log10

                      =5

For 25% completion:pH=$-logK_{a} +log(\frac{0.025}{0.075} )$

                                       =$5+log(\frac{1}{3} )$

For 50% completion:pH=$5+log(\frac{0.050}{0.050} )$

                                       =5+log1

                                       =5

For 75% completion:pH=$5+log(\frac{0.075}{0.025} )$

                                       =5+log3