Answer
The necessary volume of $HCl$ $0.105M$ is $40.2ml$
Work Step by Step
1. Calculate the molar mass $(NaOH)$:
22.99* 1 + 16* 1 + 1.01* 1 = 40.0g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 1.35}{ 40}$
$n(moles) = 0.03375$
3. Find the concentration in mol/L:
0.03375 mol in 1L = $0.03375M$
4. Find the volume necessary for an equal number of moles:
$C_1 * V_1 = C_2 * V_2$
$ 0.03375* 125= 0.105 * V_2$
$ 4.21875 = 0.105 * V_2$
$V_2 = 40.18$
