How to determine which combination of substances will give a buffered solution given target pH and Kb of each substances - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-09-21T15:26:05Z https://chemistry.stackexchange.com/feeds/question/72465 https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/72465 1 How to determine which combination of substances will give a buffered solution given target pH and Kb of each substances John Rawls https://chemistry.stackexchange.com/users/31690 2017-04-12T18:11:39Z 2017-04-12T18:25:42Z <blockquote> <p>What combination of substances will give a buffered solution that has a pH of 5.05? (Assume each pair of substances is dissolved in 5.0 L of water.) (Kb for NH3 = $1.8 × 10^{–5}$; Kb for C5H5N = $1.7 × 10^{–9}$)</p> <p>a)1.0 mole NH3 and 1.5 mole NH4Cl</p> <p>b)1.5 mole NH3 and 1.0 mole NH4Cl</p> <p>c)1.0 mole C5H5N and 1.5 mole C5H5NHCl</p> <p>d)1.5 mole C5H5N and 1.0 mole C5H5NHCl</p> </blockquote> <p>According to my text book it is choice C but I do not understand why and here was my thought process:</p> <hr> <p>For starters I change the $K_b$ into $K_a$ to get : $5.6*10^{-10}$ for Ammonia and for the other I got: $5.9*10^{-6}$</p> <p>It seems to me that this has to do with the Henderson Hasselbach Equation so I set up the following:</p> <p>$5.05=5.9*10^{-6}+log{\frac{Base}{acid}}$</p> <p>$5.05=5.9*10^{-10}+log{\frac{Base}{acid}}$</p> <p>because I already have two substances.</p> <p>But from here I feel like that wasn't the best route I should take because I got stuck right here and would like some assistance.</p> https://chemistry.stackexchange.com/questions/72465/-/72467#72467 1 Answer by Cyclohexanol. for How to determine which combination of substances will give a buffered solution given target pH and Kb of each substances Cyclohexanol. https://chemistry.stackexchange.com/users/38684 2017-04-12T18:23:05Z 2017-04-12T18:23:05Z <p>Using the Henderson-Hesselbach equation is a good idea. As you want the buffer's pH to be acidic, then using the $\ce{NH3/NH4+}$ buffer is not a good idea, as that would be basic, so we should use the $\ce{C5H5N/C5H5NHCl}$. Then, by Henderson-Hesselbach, we have that $\ce{5.05 = pH = pKa + \log \left(\frac{[C5H5N]}{[C5H5NH+]}\right)}\implies \ce{5.05 - pKa = 5.05 - 5.23 = \log \left(\frac{[C5H5N]}{[C5H5NH+]}\right)} \implies \ce{\frac{[C5H5N]}{[C5H5NH+]} = 10^{-0.18}\implies \frac{[C5H5N]}{[C5H5NH+]} \approx 0.66\approx \frac 23}.$ Thus, we want a buffer solution where the ratio of the base to the acid is approximately 2/3, which answer choice C satisfies.</p>