Answer
There are necessary 28.8 mL of that $0.100M$ $HCl$ solution to neutralize that sample of this $RbOH$ solution.
Work Step by Step
At the equivalence point, the $RbOH$ is completely neutralized by the added $HCl$.
This neutralization reaction can be represented by this balanced equation:
$RbOH(aq) + HCl(aq) -- \gt RbCl(aq) + H_2O(l)$
Therefore, to neutralizate each $RbOH$ mol, we need 1 mol of $HCl$.
- The concentration of the $RbOH$ solution is equal to $0.115M$.
- The concentration of the $HCl$ solution is equal to $0.100M$.
Use these informations as conversion factors, to calculate the necessary volume of that $HCl$ solution to neutralize 25.0 mL of that $RbOH$ solution.
$25.0mL (Soln. RbOH) \times \frac{1L}{1000mL} \times \frac{0.115 mol (RbOH)}{1L(Soln. RbOH)} \times \frac{1mol(HCl)}{1mol(RbOH)} \times \frac{1L(Soln. HCl)}{0.100mol(HCl)} \times \frac{1000mL}{1L} = 28.8mL (Soln. HCl)$
