# Question 3b25a

Dec 2, 2016

$\textsf{\left[{H}^{+}\right] = 5.34 \times {10}^{- 12} \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

$\textsf{N {H}_{3} + {H}_{2} O r i g h t \le f t h a r p \infty n s N {H}_{4}^{+} + O {H}^{-}}$

For a weak base like this we can use the expression:

sf(pOH=1/2(pK_b-log[base])

Putting in the numbers:

sf(pOH=1/2(5.00-log(0.35))#

$\therefore$$\textsf{p O H = \frac{1}{2} \times 5.4559 = 2.727}$

$\textsf{p O H + p H = 14}$

$\therefore$$\textsf{p H = 14 - p O H = 14 - 2.727 = 11.27}$

Since $\textsf{\left[{H}^{+}\right] = {10}^{- p H}}$

This gives:

$\textsf{\left[{H}^{+}\right] = 5.34 \times {10}^{- 12} \textcolor{w h i t e}{x} \text{mol/l}}$