Answer
$pH = 11.981 $
Work Step by Step
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [HB^+] = x$
-$[B] = [B]_{initial} - x = 0.61 - x$
For approximation, we consider: $[B] = 0.61M$
2. Now, use the Kb value and equation to find the 'x' value.
$Ka = \frac{[OH^-][HB^+]}{ [B]}$
$Ka = 1.5 \times 10^{- 4}= \frac{x * x}{ 6.1 \times 10^{- 1}}$
$Ka = 1.5 \times 10^{- 4}= \frac{x^2}{ 6.1 \times 10^{- 1}}$
$ 9.15 \times 10^{- 5} = x^2$
$x = 9.566 \times 10^{- 3}$
Percent ionization: $\frac{ 9.566 \times 10^{- 3}}{ 6.1 \times 10^{- 1}} \times 100\% = 1.568\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = x = 9.566 \times 10^{- 3} $
$pOH = -log[OH^-]$
$pOH = -log( 9.566 \times 10^{- 3})$
$pOH = 2.019$
$pH + pOH = 14$
$pH + 2.019 = 14$
$pH = 11.981$