# What volume of 0.200*mol*L^-1 of NaOH(aq) would react with a mass of 0.5466*g of "KHP"?

Nov 6, 2016

By $\text{KHPh}$ we assume you means $\text{potassium hydrogen phthalate}$.

#### Explanation:

$\text{KHP}$ $=$ ${C}_{6} {H}_{4} \left(C {O}_{2} H\right) \left(C {O}_{2}^{-} {K}^{+}\right)$.

$\text{Moles of KHP}$ $=$ $\frac{0.5466 \cdot g}{204.22 \cdot g \cdot m o {l}^{-} 1} = 2.677 \times {10}^{-} 3 \cdot m o l$.

Reaction:

$N a O H \left(a q\right) + \text{KHPh"(aq)rarr"KNaPh} \left(a q\right) + {H}_{2} O \left(a q\right)$

So there is $1 : 1$ equivalence, and $\text{moles of KHP "-=" moles of sodium hydroxide}$.

And thus, if $0.200 \cdot m o l \cdot {L}^{-} 1$ $\text{sodium hydroxide solution}$ is available, we need a volume of,

$\frac{2.677 \times {10}^{-} 3 \cdot \cancel{m o l}}{0.200 \cdot \cancel{m o l} \cdot \cancel{{L}^{-} 1}} \times 1000 \cdot m L \cdot \cancel{{L}^{-} 1} \cong 13 \cdot m L .$