Answer:
pH=11.
Explanation:
Hello!
In this case, since the data is not given, it is possible to use a similar problem like:
"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"
Thus, for the reaction:
[tex]C_2H_5NH_2+H^+\rightleftharpoons C_2H_5NH_4^+[/tex]
Tt is possible to compute the remaining moles of ethylamine via the following subtraction:
[tex]n_{ethylamine}=0.1850L*0.7500mol/L=0.1365mol\\\\n_{acid}=0.1144L*0.4800mol/L=0.0549mol\\\\n_{ethylamine}^{remaining}=0.1365mol-0.0549mol=0.0816mol[/tex]
Thus, the concentration of ethylamine in solution is:
[tex][ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M[/tex]
Now, we can also infer that some salt is formed, and has the following concentration:
[tex][salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M[/tex]
Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:
[tex]pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0[/tex]
Finally, the pH turns out to be:
[tex]pH=14-pOH=14-3\\\\pH=11[/tex]
NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.
Best regards!
