Question

Papaverine hydrochloride (abbreviated papH⁺Cl^-; molar mass = 378.85 g/mol) is a drug that belongs to a group of medicines called vasodilators, which cause blood vesels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; K_b = 8.33 × 10^-9 at 35.0°C). Calculate the pH of a 30.0-mg/mL aqueous dose of papH⁺Cl^- prepared at 35.0°C. Kw at 35.0°C is 2.1 × 10^-14.

Short answer

Answer

pH of the given solution comes out to be 3.35.

Step-by-step solution

 Calculation of Ka for Conjugate Base Hydrolysis

As Papaverine hydrochloride contains the weak base so the base will be hydrolysed as:

$pap{H}^{+}+H_{2}O\rightleftarrows papHOH+H^{+}$

As K_b and K_w at 35°C are given, Ka is calculated using formula

$\begin{array}{rcl}{K}_a& =& \frac{K_w}{K_b}\\ K_a& =& \frac{2.1\times {10}^{-14}}{8.33\times {10}^{-9}}\\ K_a& =& 2.52\times {10}^{-6}\end{array}$

Calculation of pH  for Given Salt

30 mg / mL means 30 g/L.

$$\begin{gathered} Conc.=\frac{g/L}{Molarmass} \\ Conc.=\frac{30}{378.85} \\ Conc.=7.92\times {10}^{-2}Mol/L \\ pap{H}^{+}+H_{2}O\rightleftarrows papHOH+H^{+} \\ 7.92\times {10}^{-2}00 \\ 7.92\times {10}^{-2}-xxx \end{gathered}$$

role="math" localid="1657864153382" $\begin{array}{rcl}{K}_a& =& \frac{{\left[x\right]}^2}{\left[7.92\times {10}^{-2}-x\right]}\\ K_a& =& \frac{x^2}{\left[7.92\times {10}^{-2}-x\right]}\\ K_a& =& 2.52\times {10}^{-6}\\ x& =& 4.47\times {10}^{-4}\\ & & \\ pH& =& -\log \left[H^{+}\right]\\ pH& =& -\log (4.47\times {10}^{-4})\\ pH& =& 3.35\end{array}$