How to find pH in the following titration/buffer problem given concentration of base and acid - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-08-19T02:02:01Z https://chemistry.stackexchange.com/feeds/question/72472 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://chemistry.stackexchange.com/q/72472 1 How to find pH in the following titration/buffer problem given concentration of base and acid John Rawls https://chemistry.stackexchange.com/users/31690 2017-04-12T19:12:36Z 2017-04-13T18:52:33Z <blockquote> <p>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 $\ce{NH3} = 1.8 × 10–5$)</p> </blockquote> <p>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:</p> <hr> <p>I first make the chemical equation:</p> <p>$\ce{NH3 + H2O &lt;--&gt;OH- + NH4+}$</p> <p>then I find that there are 0.0055 moles of NH3 and I set up the Kb problem:</p> <p>$\mathrm{Kb}=\frac{X^2}{0.0055-x}$</p> <p>then I find $x$ to be $3.15*10^{-4}$</p> <p>then I subtract that with the moles of $\ce{H+}$ ions from the $\ce{HCl}$ and divide that with $.15$ liters to get $.065$</p> <p>when I take the negative log of that I get $1.189$ but that is none of the answers.</p> <p>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. </p> <p>*I do understand that we can use the Henderson Hasselbach but I was wondering how to solve this without the equation?</p> https://chemistry.stackexchange.com/questions/72472/-/72504#72504 3 Answer by Jacie for How to find pH in the following titration/buffer problem given concentration of base and acid Jacie https://chemistry.stackexchange.com/users/43871 2017-04-13T13:28:30Z 2017-04-13T15:26:10Z <p>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 $\ce{[H]}$.</p>