If 25.21 mL of NaOH solution is required to react completely with .550 g KHP, what is the molarity of the NaOH solution?

May 14, 2016

Rxn: ${C}_{6} {H}_{4} \left(C {O}_{2} H\right) \left(C {O}_{2}^{-} {K}^{+}\right) + N a O H \left(a q\right) \rightarrow {C}_{6} {H}_{4} \left(C {O}_{2}^{-} N {a}^{+}\right) \left(C {O}_{2}^{-} {K}^{+}\right) + {H}_{2} O$

Explanation:

Given the equation, there is 1:1 equivalence between moles of KHP and moles of sodium hydroxide.

$\text{Moles of KHP}$ $=$ $\frac{0.550 \cdot g}{204.22 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.69 \times {10}^{-} 3 \cdot m o l$

$\text{Concentration of NaOH(aq) solution:}$ $=$ $\frac{2.69 \times {10}^{-} 3 \cdot m o l}{25.21 \times {10}^{-} 3 \cdot L}$ $\cong$ $0.100 \cdot m o l \cdot {L}^{-} 1$, but do the calculation for an exact value.

$\text{KHP}$ is an excellen primary standard, because we could also have used it as a base and not an acid, and used it to standardize a hydrochloirc acid solution.