# What is the pH of a solution comprising 100*mL volumes EACH of HCl(aq) at 0.20*mol*L^-1 and NaOH(aq) at 0.18*mol*L^-1?

Sep 25, 2017

$p H \equiv 2$

#### Explanation:

We interrogate the chemical reaction....

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

$\text{Moles of HCl(aq)} = 100 \times {10}^{-} 3 \cdot L \times 0.2 \cdot m o l \cdot {L}^{-} 1 = 2.0 \times {10}^{-} 2 \cdot m o l$.

$\text{Moles of NaOH(aq)} = 100 \times {10}^{-} 3 \cdot L \times 0.18 \cdot m o l \cdot {L}^{-} 1 = 1.8 \times {10}^{-} 2 \cdot m o l$.

There is thus an excess of hydronium ions, to the tune of $2.0 \times {10}^{-} 2 \cdot m o l - 1.8 \times {10}^{-} 2 \cdot m o l = 2.0 \times {10}^{-} 3 \cdot m o l$

But this is dissolved in a solution whose volume is now $200 \times {10}^{-} 3 \cdot L$

And so $\left[H C l\right] = \frac{2.0 \times {10}^{-} 3 \cdot m o l}{200 \times {10}^{-} 3 \cdot L} = 0.01 \cdot m o l \cdot {L}^{-} 1$

But $p H \equiv - {\log}_{10} \left\{\left[{H}_{3} {O}^{+}\right]\right\} = - {\log}_{10} \left(0.01\right) = 2.0$