Answer to Question #92444 in Chemistry for Tj

Question

A 25.00 mL sample of 0.379 M nitric acid (HNO3(aq)) is titrated with 12.00 mL of 0.573 M lithium hydroxide (LiOH(aq)). Determine the amount of [H+(aq)] remaining in the flask and pH of the solution after reaction

Answer 1

$HNO_3(aq) + LiOH(aq)\rightarrow LiNO_3(aq) + H_2O(l)$

1) $n=c\times V$

$n(HNO_3) = 0.379 \frac{mol}{L} \times 25.00\times 10^{-3}L = 9.475\times10^{-3}mol$

$n(LiOH) = 0.573 \frac{mol}{L}\times12.00\times 10^{-3} L = 6.876\times 10^{-3}mol$

2) As mole ratio $n(HNO_3):n(LiOH) = 1:1$ , then $n(HNO_3)$ should be equal to $n(LiOH)$

We have $n(HNO_3)>n(LiOH)$,

then LiOH is a limiting reactant.

3) $n(HNO_3)$ consumed in the reaction is equal to $n(LiOH)$

$n(HNO_3)_{consumed} = 6.976\times10^{-3}mol$

$n(HNO_3)_{left} = 9.475\times10^{-3} - 6.876\times10^{-3} = 2.599\times 10 ^{-3}mol$

4) $HNO_3(aq)\rightarrow H^+(aq) + NO_3^-(aq)$

as 1 mole of $HNO_3$ gives 1 mole of $H^+$ then $n(H^+) = n(HNO_3)_{left} =2.599\times10^{-3} mol$

$c(H^+)=\frac{n(H^+)}{V} = \frac{2.599\times 10^{-3}mol}{25.00\times10^{-3}+12\times10^{-3}} = 7.024\times10^{-2} \frac{mol}{L}$

$pH =-log_{10}[H^+] -log_{10}7.024\times10^{-2} = 1.153$