buffer solutions and ph calculations\nyou will first need to convert both the base and the acid to moles…

buffer solutions and ph calculations\nyou will first need to convert both the base and the acid to moles using the equation vxm (convert ml to l divide by 1000). then, use the henderson - hasselbalch equation to solve the following problems. the pkas have been provided for you.\n mathrm{ph}=mathrm{p}k_{a}+logleft(\frac{a^{-}}{ha}\right) \na - is base, ha is acid\n1. a student prepares a buffer solution by mixing 50.0 ml of 0.40 m acetic acid ((mathrm{ch}_{3}mathrm{cooh})) with 150.0 ml of 0.30 m sodium acetate ((mathrm{ch}_{3}mathrm{coo}^{-})). the pka of acetic acid is 4.76. what is the ph of the resulting solution?

buffer solutions and ph calculations\nyou will first need to convert both the base and the acid to moles using the equation vxm (convert ml to l divide by 1000). then, use the henderson - hasselbalch equation to solve the following problems. the pkas have been provided for you.\n mathrm{ph}=mathrm{p}k_{a}+logleft(\frac{a^{-}}{ha}\right) \na - is base, ha is acid\n1. a student prepares a buffer solution by mixing 50.0 ml of 0.40 m acetic acid ((mathrm{ch}_{3}mathrm{cooh})) with 150.0 ml of 0.30 m sodium acetate ((mathrm{ch}_{3}mathrm{coo}^{-})). the pka of acetic acid is 4.76. what is the ph of the resulting solution?

Answer

Explanation:

Step1: Calculate moles of acid

First, convert volume of acetic - acid to liters: $V_{HA}=50.0\ mL = 0.0500\ L$. Then use $n = M\times V$ to find moles of acetic acid ($HA$). $n_{HA}=M_{HA}\times V_{HA}=0.40\ mol/L\times0.0500\ L = 0.020\ mol$.

Step2: Calculate moles of base

Convert volume of sodium - acetate to liters: $V_{A^-}=150.0\ mL = 0.1500\ L$. Then use $n = M\times V$ to find moles of sodium acetate ($A^-$). $n_{A^-}=M_{A^-}\times V_{A^-}=0.30\ mol/L\times0.1500\ L = 0.045\ mol$.

Step3: Use Henderson - Hasselbalch equation

The Henderson - Hasselbalch equation is $pH = pK_a+\log\left(\frac{[A^-]}{[HA]}\right)$. In a buffer solution, when volumes are additive, the ratio of concentrations $\frac{[A^-]}{[HA]}$ is equal to the ratio of moles $\frac{n_{A^-}}{n_{HA}}$. Substitute $pK_a = 4.76$, $n_{A^-}=0.045\ mol$, and $n_{HA}=0.020\ mol$ into the equation: $pH = 4.76+\log\left(\frac{0.045}{0.020}\right)$. First, calculate $\frac{0.045}{0.020}=2.25$. Then, $\log(2.25)\approx0.35$. $pH = 4.76 + 0.35=5.11$.

Answer:

$5.11$