# What is the molarity of a NaOH solution if 28.50 mL are needed to titrated a 0.7154 g sample of KHP?

$1 , 2 - {C}_{6} {H}_{4} \left(C {O}_{2} H\right) \left(C {O}_{2}^{-} {K}^{+}\right) + N a O H \left(a q\right) \rightarrow {C}_{6} {H}_{4} \left(C {O}_{2}^{-} N {a}^{+}\right) \left(C {O}_{2}^{-} {K}^{+}\right) + {H}_{2} O \left(l\right)$
And thus the moles of $K H P$ is equivalent to the moles of sodium hydroxide, the which are dissolved in a volume of $28.50 \cdot m o l$.
$\frac{\frac{0.7154 \cdot g}{204.22 \cdot g \cdot m o {l}^{-} 1}}{28.50 \times {10}^{-} 3 \cdot L} = 0.1229 \cdot m o l \cdot {L}^{-} 1$
..........WITH RESPECT TO $N a O H \left(a q\right)$....