# A 35.0 mL sample of 0.225 M HBr was titrated with 42.3 mL of KOH. What is the concentration of the KOH?

Apr 26, 2016

Approx. $0.19 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

$H B r \left(a q\right) + K O H \left(a q\right) \rightarrow K B r \left(a q\right) + {H}_{2} O \left(l\right)$

Clearly there is a $1 : 1$ equivalence between acid and base.

$\text{Moles of HBr} = 35.0 \times {10}^{-} 3 L \times 0.225 \cdot m o l \cdot {L}^{-} 1 = 7.88 \times {10}^{-} 3 \cdot m o l .$

$\left[K O H\right]$ $=$ $\text{Moles of base"/"volume of solution}$

$=$ $\frac{7.88 \times {10}^{-} 3 \cdot m o l}{42.3 \times {10}^{-} 3 \cdot L}$ $=$ ??*mol*L^-1