Question #92444
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
1
Expert's answer
2019-08-26T05:41:50-0400

HNO3(aq)+LiOH(aq)LiNO3(aq)+H2O(l)HNO_3(aq) + LiOH(aq)\rightarrow LiNO_3(aq) + H_2O(l)

1) n=c×Vn=c\times V

n(HNO3)=0.379molL×25.00×103L=9.475×103moln(HNO_3) = 0.379 \frac{mol}{L} \times 25.00\times 10^{-3}L = 9.475\times10^{-3}mol


n(LiOH)=0.573molL×12.00×103L=6.876×103moln(LiOH) = 0.573 \frac{mol}{L}\times12.00\times 10^{-3} L = 6.876\times 10^{-3}mol


2) As mole ratio n(HNO3):n(LiOH)=1:1n(HNO_3):n(LiOH) = 1:1 , then n(HNO3)n(HNO_3) should be equal to n(LiOH)n(LiOH)

We have n(HNO3)>n(LiOH)n(HNO_3)>n(LiOH),

then LiOH is a limiting reactant.


3) n(HNO3)n(HNO_3) consumed in the reaction is equal to n(LiOH)n(LiOH)

n(HNO3)consumed=6.976×103moln(HNO_3)_{consumed} = 6.976\times10^{-3}mol

n(HNO3)left=9.475×1036.876×103=2.599×103moln(HNO_3)_{left} = 9.475\times10^{-3} - 6.876\times10^{-3} = 2.599\times 10 ^{-3}mol


4) HNO3(aq)H+(aq)+NO3(aq)HNO_3(aq)\rightarrow H^+(aq) + NO_3^-(aq)

as 1 mole of HNO3HNO_3 gives 1 mole of H+H^+ then n(H+)=n(HNO3)left=2.599×103moln(H^+) = n(HNO_3)_{left} =2.599\times10^{-3} mol

c(H+)=n(H+)V=2.599×103mol25.00×103+12×103=7.024×102molLc(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=log10[H+]log107.024×102=1.153pH =-log_{10}[H^+] -log_{10}7.024\times10^{-2} = 1.153


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