How many milliliters of .45 M #HCl# will neutralize 25.0 mL of 1.00 M #KOH#?

Question

Caleb Adams · Accepted Answer

Approx. #60*mL# of hydrochloric acid are required.

For every problem of this kind, the equivalency must be established using a BALANCED chemical equation: #KOH(aq) + HCl(aq) rarr KCl(aq) + H_2O(l)# In light of the 1:1 equivalency, #"Moles of KOH"=25xx10^-3*Lxx1.00*mol*L^-1=25xx10^-3*mol# Thus we require an equivalent quantity of #0.45*mol*L^-1# hydrochloric acid: #=# #(25xx10^-3*mol)/(0.45*mol*L^-1)xx10^3*mL*L^-1# #=# #??mL#

Wyatt Atkins · Answer

undefined To find the volume of HCl needed to neutralize the KOH:

1. Determine the moles of KOH: $1.00 \, \text{M} \times 25.0 \, \text{mL} = 25.0 \, \text{mmol}$.
2. According to the balanced equation for neutralization, 1 mole of KOH reacts with 1 mole of HCl.
3. Since the concentrations are given, you can use the equation $M_1V_1 = M_2V_2$ to find the volume of HCl needed:
$\frac{{M_1 \times V_1}}{{M_2}} = V_2$.
4. Substitute the given values:
$\frac{{0.45 \, \text{M} \times V_2}}{{1.00 \, \text{M}}} = 25.0 \, \text{mmol}$.
5. Solve for $V_2$:
$V_2 = \frac{{25.0 \, \text{mmol}}}{{0.45}} = 55.6 \, \text{mL}$.

Therefore, 55.6 mL of 0.45 M HCl is needed to neutralize 25.0 mL of 1.00 M KOH.