Answer
$pH = 0.824$
Work Step by Step
1. Identify all the electrolytes:
$N{H_4}^+$: Weak acid.
$Cl^-$: Insignificant base
$HNO_3$: Strong acid.
2. Find the hydronium ion concentration given by the $HNO_3$
Strong acid: $[H_3O^+] = [HNO_3] = 0.150M$
3. Now, we could calculate the hydronium ion given by the $N{H_4}^+$, but since this is a very small value, we can ignore it, and just use the $[H_3O^+] = 0.150M$
$K_a (N{H_4}^+) = \frac{10^{-14}}{ 1.76\times 10^{- 5}**} = 5.68 \times 10^{-10}$
** $K_b$ for $NH_3$
4. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.150M)$
$pH = 0.824$
