Answer
$pH = 10.2$
Work Step by Step
1. Find the number of moles of acid, and base:
n(moles) = c(mol/L) * V(L)
Acid: n = 0.0155* 0.05 = $7.75\times 10^{- 4}$moles
Base: n = 0.0106* 0.075 = $ 7.95\times 10^{- 4}$moles
2. Calculate the difference between them:
$7.95\times 10^{- 4} - 7.75\times 10^{- 4}= 2.0\times 10^{- 5} moles$
** This difference is related to the acid-base neutralization.
** The base has a higher number of moles, therefore, this is the excess of $[OH^-]$
3. Find the concentration of $[OH^-]$
- $C(mol/L) = n(moles) / V(L)$
** V(L) = 0.05 + 0.075 = 0.125L
- $C = \frac{2.0 \times 10^{-5} moles}{0.125L} = 1.6 \times 10^{-4}M $
4. Convert that number into pH.
$pOH = -log[OH^-]$
$pOH = -log( 1.6 \times 10^{- 4})$
$pOH = 3.79$
$pH + pOH = 14$
$pH + 3.79 = 14$
$pH = 10.2$
