# What is [HCl] if a 25*mL volume of [NaOH] of 0.250*mol*L^-1 concentration reaches an endpoint with a 22.5*mL volume of the acid?

Jul 8, 2017

You mean to start with a solution of $N a O H \left(a q\right)$

#### Explanation:

$N a O H \left(a q\right)$ reacts with $H C l \left(a q\right)$ with 1:1 stoichiometry.......

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

Given that $\text{Concentration"="Moles of titrant"/"Volume of titrant}$ we can work out the molar equivalence......

$\text{Moles of NaOH}$ $=$ $0.250 \cdot m o l \cdot {L}^{-} 1 \times 25 \times {10}^{-} 3 \cdot L = 6.25 \times {10}^{-} 3 \cdot m o l$, and given the 1:1 stoichiometry, this was also the number of moles of base neutralized.

This molar quantity was equivalent to the amount of moles contained in $22.5 \cdot c {m}^{3}$ $H C l$ titrant..........so..........

$\text{Concentration}$ $=$ $\frac{6.25 \times {10}^{-} 3 \cdot m o l}{22.5 \times {10}^{-} 3 \cdot L}$

$\left[H C l\right] = 0.278 \cdot m o l \cdot {L}^{-} 1$.........

Note that $1 \cdot L = 1000 \cdot c {m}^{3} = 1000 \cdot m L$...........