# In standardization, 8.77 mL of NaOH neutralized 1.522 g of KHP. Given the molar mass of KHP is 204.22 g/mol, what is the molarity of the NaOH solution?

##### 1 Answer
Jun 13, 2017

$\text{Molarity} = 0.8500 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

$K H P$, $\text{potassium hydrogen phthalate}$, $1 , 2 - {C}_{6} {H}_{4} C {O}_{2} H C {O}_{2}^{-} {K}^{+}$, is the potassium salt of the diacid, $\text{pthalic acid}$, $1 , 2 - {C}_{6} {H}_{4} {\left(C {O}_{2} H\right)}_{2}$. It can be titrated as a base, or here as an acid. It is air stable, non-hygroscopic, and thus useful as a primary standard.

$1 , 2 - {C}_{6} {H}_{4} C {O}_{2} H C {O}_{2}^{-} {K}^{+} + N {a}^{+} \text{^(-)OH rarr "1,2-} {C}_{6} {H}_{4} C {O}_{2}^{-} N {a}^{+} C {O}_{2}^{-} {K}^{+} + {H}_{2} O$

And thus moles of $N a O H$ $\equiv$ $\text{KHP}$, and thus............

$\left[N a O H\right] = \frac{\frac{1.522 \cdot g}{204.22 \cdot g \cdot m o {l}^{-} 1}}{8.77 \times {10}^{3} \cdot L} = 0.8500 \cdot m o l \cdot {L}^{-} 1$.