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A $100. \text{ mL}$ sample of $0.10 \text{M } \ce{HCl}$ is mixed with $50. \text{ mL of } 0.11 \text{M } \ce{NH3}$. What is the resulting pH? (Kb for $\ce{NH3} = 1.8 × 10–5$)

I was thinking more along the lines of trying to find out the moles of acid and the moles of bases that we have and subtracting the moles and finding the concentrations so I did the following:


I first make the chemical equation:

$\ce{NH3 + H2O <-->OH- + NH4+}$

then I find that there are 0.0055 moles of NH3 and I set up the Kb problem:

$\mathrm{Kb}=\frac{X^2}{0.0055-x}$

then I find $x$ to be $3.15*10^{-4}$

then I subtract that with the moles of $\ce{H+}$ ions from the $\ce{HCl}$ and divide that with $.15$ liters to get $.065$

when I take the negative log of that I get $1.189$ but that is none of the answers.

Why is this and what source of error have I made and is there a concept that I do not understand? Thus I request for assistance.

*I do understand that we can use the Henderson Hasselbach but I was wondering how to solve this without the equation?

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This is a titration problem - not necessarily a buffer problem. While you have $0.0055$ moles of $\ce{NH3}$, you have $0.01$ moles of $\ce{H}$ from $\ce{HCl}$ dissociating. $0.01 - 0.0055 = 0.0045$ moles of $\ce{H}$ left. Take H/total L to find M. Take the $-\log$ of $\ce{[H]}$.

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