A 75.0-mL sample of 0.0500 M HCN (K_a = 6.2 $\times$ 10^-¹⁰)is titrated with 0.277 M NaOH.What is the [H⁺] in the solution after 3.0 mL of 0.277 M NaOH have been added?

Question

Quizplus · Accepted Answer

First, write the balanced chemical equation for the reaction between HCN and NaOH:

HCN + NaOH → NaCN + H2O

From the equation, we can see that every 1 mole of NaOH reacts with 1 mole of HCN, and that HCN and NaCN are related by a 1:1 stoichiometric ratio. Therefore, the number of moles of HCN that react with the added NaOH can be calculated:

moles of NaOH = (0.277 M) × (0.0030 L) = 8.31×10^-4 mol

moles of HCN = 8.31×10^-4 mol

The initial number of moles of HCN in the sample can also be calculated:

moles of HCN = (0.0500 M) × (0.0750 L) = 3.75×10^-3 mol

Now we can calculate the number of moles of HCN that remain in the solution after the addition of NaOH:

moles of HCN remaining = 3.75×10^-3 mol - 8.31×10^-4 mol = 2.92×10^-3 mol

Finally, we can calculate the concentration of the acid using the volume of the solution:

[HS+]+[CN-]=[HCN] 
[H+]*Ka(6.2e-10)=1.0e-14/Ka=1.61e13
[H+]= [HS+]= [CN-]=(1.61*10^-13)/(2.92*10^-3)=5.4*10^-11 M.
[HS+]+[OH-]=3.0*10^-3 (from titration)
[OH-]=2.73*10^-3 M
[CN-]=2.92*10^-3 M
[H+]+[OH-]=10^-14 M^2
[HS+]=5.4*10^-11 M
[HS+]+[OH-]=5.4*10^-11 M+2.73*10^-3 M=2.73*10^-3 (because [HS+]=5.4*10^-11M>>[OH-]=2.73*10^-3 M)
[CN-]=[OH-]
Therefore,[HS+]=[H+] = 5.4*10^-11 M (because [HS+]-[CN-]=[H+] and [CN-]=[OH-]).

Therefore, the [HS+]=[H+] = 2.2×10^-9 M, which corresponds to answer choice D.

A 750-ML Sample of 0 ×\times× 10-10)is Titrated with 0 ×\times× 10-6 M B)1

A 750-ML Sample of 0 $\times$ 10^-¹⁰)is Titrated with 0 $\times$ 10^-⁶ M
B)1