# In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 mL of HCl to completely neutralize the acid?

May 16, 2017

Well, you need to add $45.0 \cdot m L$ of your standard $N a O H$ solution precisely...........

#### Explanation:

And first we represent the reaction by an equation:

$H C l \left(a q\right) + N a O H \left(a q\right) \rightarrow N a C l \left(a q\right) + {H}_{2} O \left(l\right)$

And thus there is $1 : 1$ $\text{stoichiometry}$.

We use the relationship $\text{Molarity"="Moles of solute"/"Volume of solution}$.

What is the volume of the solution at equivalence?