# What is the pH of a solution made by mixing 200 mL of 0.0657 M NaOH, 140 mL of 0.107 M HCl, and 160 mL of H2O? I know that the answer is 2.43, but could someone please explain how? Thank you so much!!!!

Dec 31, 2017

$p H \equiv - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left(3.68 \times {10}^{-} 3\right) = - \left(- 2.43\right) = 2.43$

#### Explanation:

We assess the chemical reaction....

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

We must calculate the stoichiometric equivalence of the acid and the base...

$\text{Moles of HCl} = 140 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.107 \cdot m o l \cdot {L}^{-} 1 = 0.01498 \cdot m o l .$

$\text{Moles of NaOH} = 200 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.0657 \cdot m o l \cdot {L}^{-} 1 = 0.01314 \cdot m o l .$

And thus, with respect to $H C l$ there is a concentration of....

$\frac{\left(0.01498 - 0.01314\right) \cdot m o l}{\left(140 \cdot m L + 200 \cdot m L + 160 \cdot m L\right) \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}$

And so...$\left[H C l\right] = \frac{1.84 \times {10}^{-} 3 \cdot m o l}{0.500 \cdot L} = 3.68 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$...and thus $\left[{H}_{3} {O}^{+}\right] = 3.68 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

Now.............. $p H \equiv - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left(3.68 \times {10}^{-} 3\right) = - \left(- 2.43\right) = 2.43$ .........as required...

If there is something that I have glossed over, or something that you do not follow, voice your concern, and someone will address it.... The basic definition is $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$; of course we first had to access $\left[{H}_{3} {O}^{+}\right]$.