Question

what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid requires 29.80 ml of 0.3567 m naoh? ka =1.8 × 10-4 for formic acid.

Answer 1

Answer:

pH at equivalence point is 8.52

Explanation:

$HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O$

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of $HCOO^{-}$

So, moles of NaOH used to reach equivalence point equal to number of moles $HCOO^{-}$ produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of $HCOO^{-}$ produced = $\frac{29.80\times 0.3567}{1000}moles=0.01063moles$

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of $HCOO^{-}$ = $\frac{0.01063\times 1000}{54.80}M=0.1940M$

At equivalence point, pH depends upon hydrolysis of $HCOO^{-}$. So, we have to construct an ICE table.

$HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}$

I: 0.1940                                   0                 0

C: -x                                          +x               +x

E: 0.1940-x                                x                  x

So, $\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}$

species inside third bracket represent equilibrium concentrations

So, $\frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}$

or,$x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0$

So, $x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}$

So, $x=3.285\times 10^{-6}M$

So, $pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52$