Answer
$[OH^-] = 0.01547M$
$pH = 12.189$
Work Step by Step
1. Find the number of moles of $OH^-$ in the 2 strong bases:
- 20ml = 0.020L; 40ml = 0.040L - ml to L: Divide by 1000.
n(moles) = c(mol/L) * V(L)
- Since we are looking for $[OH^-]$, we multiply these numbers by the quatity of $OH$ in each molecule.
$Ba(OH)_2$ : $n = 0.015* 0.02 * 2 = 6\times 10^{- 4}moles$
$NaOH$: $n = 0.0082* 0.04 * 1 = 3.28\times 10^{- 4}moles$
2. Sum these numbers:
$OH^- = 6\times 10^{- 4} + 3.28\times 10^{- 4}$
$OH^- = 9.28\times 10^{- 4}$
3. Find the total volume, and calculate the concentration:
- Total volume = 0.02 + 0.04 = 0.06L
$Conc(mol/L) = \frac{n(moles)}{V(L)}$
$C(mol/L) = \frac{ 9.28\times 10^{- 4}}{ 0.06}$
$C(mol/L) = 0.01547M$
4. Calculate the pOH, then the pH:
$pOH = -log[OH^-]$
$pOH = -log( 0.01547)$
$pOH = 1.811$
$pH + pOH = 14$
$pH + 1.811 = 14$
$pH = 12.189$
