## Chemistry: The Central Science (13th Edition)

$[OH^-] = 0.1128M$ $pH = 13.052$
1. Calculate the molar mass: Molar Mass ($KOH$): 39.1* 1 + 16* 1 + 1.01* 1 = 56.11g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 3.165}{ 56.11}$ $n(moles) = 0.05641$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ - 500ml = 0.5L $C(mol/L) = \frac{ 0.05641}{ 0.5}$ $C(mol/L) = 0.1128$ - Since KOH is a strong base:$[OH^-] = [KOH] = 0.1128M$ 4. Now, calculate the pOH, and then the pH: $pOH = -log[OH^-]$ $pOH = -log( 0.1128)$ $pOH = 0.943$ $pH + pOH = 14$ $pH + 0.9476 = 14$ $pH = 13.052$