# Three beakers each contain "100 mL" of acidic solution with a "pH" of 3.00. The acids in the beakers are "HCl", "HNO"_2, and "HI". If "50 mL" of "0.1 M NaOH" is added to each beaker, which resulting solution will have the lowest "pH"?

Apr 18, 2018

All three solutions have the same pH.

#### Explanation:

$\text{HCl}$ and $\text{HI}$

The $\text{HCl}$ and $\text{HI}$ are both strong acids, so they will each give the same result.

Let's call them $\text{HX}$.

If

$\text{pH = 3.00}$

Then

["H"_3"O"^"+"] = 10^"-3.00"color(white)(l)"mol/L" = 1.00 × 10^"-3"color(white)(l)"mol/L"

The acid will react completely with the $\text{NaOH}$.

$\text{Moles of HX" = 100 color(red)(cancel(color(black)("mL HX"))) × (1.00 × 10^"-3" color(white)(l)"mmol HX")/(1 color(red)(cancel(color(black)("mL HX")))) = "0.100 mmol HX}$

$\text{Moles of NaOH" = 50 color(red)(cancel(color(black)("mL NaOH"))) × "0.1 mol NaOH"/(1 color(red)(cancel(color(black)("mL NaOH")))) = "5.0 mmol NaOH}$

$\textcolor{w h i t e}{m m m m m l l} \text{HX + NaOH → NaX" + "H"_2"O}$
$\text{I/mol} : \textcolor{w h i t e}{m l l} 0.100 \textcolor{w h i t e}{m m} 5.0$
$\text{C/mol": color(white)(m)"-0.100"color(white)(ml)"-0.100}$
$\text{E/mol} : \textcolor{w h i t e}{m l l} 0 \textcolor{w h i t e}{m m m l l} 4.9$

So, we have 4.9 mmol of $\text{NaOH}$ in 150 mL of solution.

["OH"^"-"] = "4.9 mmol"/"150 mL" = "0.033 mol/L"

$\text{pOH = -log} 0.033 = 1.5$

$\text{pH = 14.00 - pOH = 14.00 - 1.5 = 12.5}$

$\boldsymbol{{\text{HNO}}_{2}}$

${\text{HNO}}_{2}$ is a weak acid with K_text(a) = 4.0 × 10^"-4".

We must calculate the initial concentration of ${\text{HNO}}_{2}$ that will give a final concentration of 1.00 × 10^"-3"color(white)(l)"mol/L H"_3"O"^"+".

$\textcolor{w h i t e}{m m m m m l m m} \text{HNO"_2 + "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) +color(white)(mm) "NO"_2^"-}$
$\text{I/mol·l"^"-1} : \textcolor{w h i t e}{m m m m} c \textcolor{w h i t e}{m m m m m m m l l} 0 \textcolor{w h i t e}{m m m m m m l l} 0$
$\text{C/mol·l"^"-1": color(white)(m)"-1.00 × 10"^"-3"color(white)(mm)"+1.00 × 10"^"-3"color(white)(m)"+1.00 × 10"^"-3}$
$\text{E/mol·l"^"-1": color(white)(m)c"-1.00 × 10"^"-3"color(white)(mml)"1.00 × 10"^"-3"color(white)(mm)"1.00 × 10"^"-3}$

K_text(a) = (["H"_3"O"^"+"]["NO"_2^"-"])/(["HNO"_2]) = (1.00 × 10^"-3")^2/(c - 1.00 ×10^"-3") = 4.0 × 10^"-4"

1.00 × 10^"-6" = 4.0 × 10^"-4"c - 4.00 × 10^"-7"

c = (1.00 × 10^"-6" + 4.00 × 10^"-7")/(4.0 × 10^"-4") = (1.40 × 10^"-6")/(4.0 × 10^"-4") = 3.5 × 10^"-3"

["HNO"_2] = 3.5 × 10^"-3"color(white)(l)"mol/L"

The ${\text{HNO}}_{2}$ will react completely with the $\text{NaOH}$.

$\text{Moles of HNO"_2 = 100 color(red)(cancel(color(black)("mL HNO"_2))) × (3.5 × 10^"-3" color(white)(l)"mmol HNO"_2)/(1 color(red)(cancel(color(black)("mL HNO"_2)))) = "0.35 mmol HX}$

$\textcolor{w h i t e}{m m m m l l} \text{HNO"_2 + "NaOH → NaX" + "H"_2"O}$
$\text{I/mol} : \textcolor{w h i t e}{m l l} 0.35 \textcolor{w h i t e}{m m m} 5.0$
$\text{C/mol": color(white)(m)"-0.35"color(white)(mml)"-0.35}$
$\text{E/mol} : \textcolor{w h i t e}{m l l} 0 \textcolor{w h i t e}{m m m m} 4.6$

So, we have 4.6 mmol of $\text{NaOH}$ in 150 mL of solution.

["OH"^"-"] = "4.6 mmol"/"150 mL" = "0.033 mol/L"

$\text{pOH = -log} 0.031 = 1.5$

$\text{pH = 14.00 - pOH = 14.00 - 1.5 = 12.5}$

All three solutions have the same pH.