Answer
$pH = 12.87$
Work Step by Step
1. Identify the significant electrolytes:
- $RbOH$: Strong base.
- $Na^+$: Insignificant.
- $HC{O_3}^-$: Weak base.
2. Calculate the $K_b$ for the weak base:
- Since $HC{O_3}^-$ is the conjugate base of $H_2CO_3$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 4.3\times 10^{- 7} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 4.3\times 10^{- 7}}$
$K_b = 2.326\times 10^{- 8}$
3. Since it has a low $K_b$ value, we can ignore it, and calculate the pH just considering $RbOH$ $0.075M$
$pOH = -log[OH^-]$
$pOH = -log( 0.075)$
$pOH = 1.13$
$pH + pOH = 14$
$pH + 1.13 = 14$
$pH = 12.87$
