## General Chemistry: Principles and Modern Applications (10th Edition)

$pH = 10.2$
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. $pOH = -log[OH^-]$ $pOH = -log( 1.6 \times 10^{- 4})$ $pOH = 3.79$ $pH + pOH = 14$ $pH + 3.79 = 14$ $pH = 10.2$