Chapter 16, Problem 16.123QA

Interpretation Introduction

To calculate:

The $pH$ at equivalence point for the titration of formic acid and $NaOH$ and boric acid and $NaOH$. Also explain if the same indicator be used for both titrations.

Answer to Problem 16.123QA

Solution:

a) pH at the equivalence point of titration of formic acid is 8.23.

b) pH at the equivalence point of titration of boric acid is 10.98.

c) Considering the difference in the pH values at equivalence point, the same indicator should not be used for both titrations.

Explanation of Solution

1) Concept:

We are asked to find the =$pH$ at the equivalence point for the titration of known concentrations of formic acid and $NaOH$ and boric acid and $NaOH$. Also, we are asked if we could use the same indicator for both of these titrations.

Formic acid is a weak organic monoprotic acid with the $K_a=1.77\times {10}^{-4},$ and $NaOH$ is a strong base. So, at the equivalence point, the salt which is formed, sodium formate, should have the basic $pH,$ that is,$pH>7$.

Boric acid is a weak monobasic acid. For the titration between boric acid and $NaOH$, the salt that is formed during the reaction is basic salt, and so, it should have the basic $pH$, that is,$pH>7$.

2) Formulae:

i) $n=V\times M$

ii) $K_a\times K_b=K_w=1 \times {10}^{-14}$

iii) $p{K}_a= -\log K_a$

iv) $pOH= -\mathrm{log }\left[O{H}^{-}\right]$

v) $K_{ }=\frac{\left[products\right]}{\left[reactants\right]}$

vi) $pH+pOH=14$

3) Given:

i) Volume of formic acid $=10.0 mL$

ii) Molarity of formic acid $=0.100 M$

iii) $K_a$ of formic acid $=1.77\times {10}^{-4}$

iv) Volume of boric acid $=10.0 mL$

v) Molarity of boric acid $=0.100 M$

vi) Molarity of $NaOH=0.100M$

vii) $K_{a1}$ of boric acid $=5.4\times {10}^{-10}$

4) Calculations:

a) Calculating the $pH$ at equivalence point for the titration of formic acid and $NaOH:$

Calculating moles of formic acid using given molarity and volume:

$n_{formic acid}=V_{formic acid}\times M_{formic acid}$

$n_{formic acid}=0.010L \times 0.100 M=0.001 mol$

At the equivalence point, moles of formic acid and sodium hydroxide are equal. Thus, moles of $NaOH$ at the equivalence point $=0.001 mol$

Calculating the volume of $NaOH$ at the equivalence point using the given molarity and calculated moles:

$n=V\times M$

$0.001=V\times 0.1 M$

$V=\frac{0.001}{0.1}=0.01 L=10 mL$

Total volume of solution is $20 mL=0.02 L$

At the equivalence point, moles of sodium formate formed are $0.001 mol$. Calculating molarity of sodium formate:

$M=\frac{n}{V}=\frac{0.001M}{0.02 L}=0.05 M$

At the equivalence point, salt sodium formate $\left(HCOONa\right)$ will form, for which the dissociation reaction is written as

 $HCOONa\left(aq\right)+H_{2}O\left(aq\right)\rightleftharpoons HCOOH \left(aq\right)+O{H}^{-}\left(aq\right)$

$K_a$ for the formic acid is $1.77\times {10}^{-4}$; the $K_b$ for sodium formate is

$K_a\times K_b=1 \times {10}^{-14}$

$1.77\times {10}^{-4}\times K_b=1 \times {10}^{-14}$

$K_b=\frac{1 \times {10}^{-14}}{1.77\times {10}^{-4}}=5.650 \times {10}^{-11}$

Creating an ICE table for the dissociation reaction of sodium formate as

$HCOONa\left(aq\right)+H_{2}O\left(aq\right) \rightleftharpoons HCOOH \left(aq\right) + O{H}^{-}\left(aq\right)$
Initial	$0.05$	$0.0$	$0.0$
Change	$-x$	$+x$	$+x$
Equilibrium	$(0.05-x)$	$x$	$x$

Writing the $K_{b1}$ and solving it for $O{H}^{-}$ ions as

$K_{b1}= \frac{{\left(x\right)}^2}{(0.05-x)}=5.650 \times {10}^{-11}$

$x^2= 5.650 \times {10}^{-11}\left(0.05\right)=2.825 \times {10}^{-12}$

$x= \sqrt{2.825 \times {10}^{-12}}=1.681 \times {10}^{-6}=\left[O{H}^{-}\right]$

$pOH= -\log 1.681 \times {10}^{-6}$$= 5.774$

$pH=14-5.774=8.23$

The $pH$ at the equivalence point for the titration between formic acid and sodium hydroxide is $8.23$.

Indicator changes color over a $pH$ range extending $\pm 1$ unit on either sides of its $p{K}_a$. The $p{K}_a$ value for phenolphthalein is close to around $8.9$ (referring figure 16.5), which is close to the $pH$ at equivalence point $\left(8.22\right)$. So, phenolphthalein should be used for the titration of formic acid and sodium hydroxide.

b) Calculating the $pH$ at equivalence point for the titration of boric acid and $NaOH$:

Calculating moles of boric acid using given molarity and volume. Here, we assume that the volume of boric acid is $10.0 mL$

$n_{boric acid}=V_{boric acid}\times M_{boric acid}$

$n_{boric acid}=0.010L \times 0.100 M=0.001 mol$

At the equivalence point, moles of formic acid and sodium hydroxide are equal. Thus, moles of $NaOH$ at the equivalence point $=0.001 mol$.

Calculating volume of $NaOH$ at the equivalence point using the given molarity and calculated moles:

$n=V\times M$

$0.001=V\times 0.1 M$

$V=\frac{0.001}{0.1}=0.01 L=10 mL$

Total volume of solution is $20 mL=0.02 L$

During the reaction of boric acid and sodium hydroxide, sodium borate will be produced. At the equivalence point, moles of sodium borate formed are $0.001 mol$, calculating the molarity of sodium borate:

$M=\frac{n}{V}=\frac{0.001M}{0.02 L}=0.05 M$

At the equivalence point, salt sodium borate $\left(B{\left(OH\right)}_{2}ONa\right)$ will form, for which the dissociation reaction is written as

 $B{\left(OH\right)}_{2}ONa\left(aq\right)+H_{2}O\left(aq\right)\rightleftharpoons B{\left(OH\right)}_3 \left(aq\right)+O{H}^{-}\left(aq\right)$

$K_{a1}$ for the boric acid is $5.4\times {10}^{-10}$. The $K_{b1}$ for sodium formate is

$K_{a1}\times K_{b1}=1 \times {10}^{-14}$

$5.4\times {10}^{-10}\times K_b=1 \times {10}^{-14}$

$5.4\times {10}^{-10}\times K_b=1 \times {10}^{-14}$

$K_{b1}=\frac{1 \times {10}^{-14}}{5.4\times {10}^{-10}}=1.852 \times {10}^{-5}$

Creating an RICE table for the dissociation reaction of sodium borate as:

$B{\left(OH\right)}_{3}ONa\left(aq\right)+H_{2}O\left(aq\right) \rightleftharpoons B{\left(OH\right)}_3 \left(aq\right) + O{H}^{-}\left(aq\right)$
Initial	$0.05$	$0.0$	$0.0$
Change	$-x$	$+x$	$+x$
Equilibrium	$(0.05-x)$	$x$	$x$

Writing the $K_{b1}$ and solving it for $O{H}^{-}$ ions is

$K_{b1}= \frac{{\left(x\right)}^2}{(0.05-x)}=1.852 \times {10}^{-5}$

$x^2= 1.852 \times {10}^{-5} \left(0.05\right)=9.26 \times {10}^{-7}$

$x= \sqrt{9.26 \times {10}^{-7}}=9.623 \times {10}^{-4}=\left[O{H}^{-}\right]$

$pOH= -\mathrm{log }9.623 \times {10}^{-4}= 3.017$

$pH=14-3.017=10.98$

The $pH$ at the equivalence point for the titration between boric acid and sodium hydroxide is $10.98$, which is in the basic range that we have predicted.

Indicator changes color over a $pH$ range extending $\pm 1$ unit on either sides of its $p{K}_a$. The $p{K}_a$ value for Alizarin Yellow R is close to around $11$ (referring figure 16.5), which is close to the $pH$ at equivalence point $\left(10.98\right)$. So, Alizarin Yellow R should be used for the titration of formic acid and sodium hydroxide.

Conclusion:

Indicator changes color over a $pH$ range extending $\pm 1$ unit on either sides of its $p{K}_a$. So, to select an indicator, the $p{K}_a$ of the indicator must be close to the $pH$ at equivalence point in the titration.