Answer
There are necessary 30.6 mL of that $0.200M$ $KOH$ solution to neutralize that sample of this $HBr$ solution.
Work Step by Step
At the equivalence point, the $HBr$ is completely neutralized by the added $KOH$.
This neutralization reaction can be represented by this balanced equation:
$HBr(aq) + KOH(aq) -- \gt KBr(aq) + H_2O(l)$
Therefore, to neutralizate each $HBr$ mol, we need 1 mol of $KOH$.
- The concentration of the $HBr$ solution is equal to $0.175M$.
- The concentration of the $KOH$ solution is equal to $0.200M$.
Use these informations as conversion factors, to calculate the necessary volume of that $KOH$ solution, to neutralize 35.0 mL of that $HBr$ solution.
$35.0mL (Soln. HBr) \times \frac{1L}{1000mL} \times \frac{0.175 mol (HBr)}{1L(Soln. HBr)} \times \frac{1mol(KOH)}{1mol(HBr)} \times \frac{1L(Soln. KOH)}{0.200mol(KOH)} \times \frac{1000mL}{1L} = 30.6mL (Soln. KOH)$
