Question

iclicker question
consider a solution containing 0.50 mol of co₃²⁻. determine the ph of the solution after 0.20 mol of hcl were added. pkₐ(hco₃⁻) = 10.33
a. 10.51
b. 10.73
c. 10.33
d. 10.15
e. 9.93

Explanation:

Step1: Write the reaction equation

$$CO_{3}^{2 - }+HCl ightarrow HCO_{3}^{-}+Cl^{-}$$

The initial moles of \(CO_{3}^{2 - } = 0.50\space mol\), moles of \(HCl=0.20\space mol\). After the reaction, moles of \(CO_{3}^{2 - }\) left \(n_{CO_{3}^{2 - }}=(0.50 - 0.20)\space mol = 0.30\space mol\), moles of \(HCO_{3}^{-}\) formed \(n_{HCO_{3}^{-}}=0.20\space mol\)

Step2: Use the Henderson - Hasselbalch equation

The Henderson - Hasselbalch equation is \(pH=pK_{a}+\log\frac{[A^{-}]}{[HA]}\). Here, \(A^{-}\) is \(CO_{3}^{2 - }\) and \(HA\) is \(HCO_{3}^{-}\), and since the volume is the same for both species (they are in the same solution), we can use the mole ratio instead of concentration ratio.
Substitute \(pK_{a}=10.33\), \([A^{-}]=0.30\space mol\), \([HA]=0.20\space mol\) into the equation:
\(pH = 10.33+\log\frac{0.30}{0.20}\)
\(\log\frac{0.30}{0.20}=\log(1.5)\approx0.18\)
\(pH=10.33 + 0.18=10.51\)

Answer:

A. 10.51