Three beakers each contain $100 mL$ $"100 mL"$ of acidic solution with a $pH$ $"pH"$ of $3.00$ $3.00$ . The acids in the beakers are $HCl$ $"HCl"$ , ${HNO}_{2}$ $"HNO"_2$ , and $HI$ $"HI"$ . If $50 mL$ $"50 mL"$ of $0.1 M NaOH$ $"0.1 M NaOH"$ is added to each beaker, which resulting solution will have the lowest $pH$ $"pH"$ ?

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Three beakers each contain $100 mL$ $"100 mL"$ of acidic solution with a $pH$ $"pH"$ of $3.00$ $3.00$ . The acids in the beakers are $HCl$ $"HCl"$ , ${HNO}_{2}$ $"HNO"_2$ , and $HI$ $"HI"$ . If $50 mL$ $"50 mL"$ of $0.1 M NaOH$ $"0.1 M NaOH"$ is added to each beaker, which resulting solution will have the lowest $pH$ $"pH"$ ?

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Answer 1 · 2018-04-18T11:15:21

All three solutions have the same pH.

Explanation:

$HCl$ $"HCl"$ and $HI$ $"HI"$

The $HCl$ $"HCl"$ and $HI$ $"HI"$ are both strong acids, so they will each give the same result.

Let's call them $HX$ $"HX"$ .

If

$pH = 3.00$ $"pH = 3.00"$

Then

$[H_{3} O^{+}] = 10^{-3.00} l mol/L = 1.00 \times 10^{-3} l mol/L$ $["H"_3"O"^"+"] = 10^"-3.00"color(white)(l)"mol/L" = 1.00 × 10^"-3"color(white)(l)"mol/L"$

The acid will react completely with the $NaOH$ $"NaOH"$ .

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

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

$color(white)(mmmmmll)"HX + NaOH → NaX" + "H"_2"O"$
$"I/mol": color(white)(mll)0.100 color(white)(mm)5.0$
$"C/mol": color(white)(m)"-0.100"color(white)(ml)"-0.100"$
$"E/mol": color(white)(mll)0color(white)(mmmll)4.9$

So, we have 4.9 mmol of $"NaOH"$ in 150 mL of solution.

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

$"pOH = -log"0.033 = 1.5$

$"pH = 14.00 - pOH = 14.00 - 1.5 = 12.5"$

$bb("HNO"_2)$

$"HNO"_2$ is a weak acid with $K_text(a) = 4.0 × 10^"-4"$ .

We must calculate the initial concentration of $"HNO"_2$ that will give a final concentration of $1.00 × 10^"-3"color(white)(l)"mol/L H"_3"O"^"+"$ .

$color(white)(mmmmmlmm)"HNO"_2 + "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) +color(white)(mm) "NO"_2^"-"$
$"I/mol·l"^"-1": color(white)(mmmm)c color(white)(mmmmmmmll)0color(white)(mmmmmmll)0$
$"C/mol·l"^"-1": color(white)(m)"-1.00 × 10"^"-3"color(white)(mm)"+1.00 × 10"^"-3"color(white)(m)"+1.00 × 10"^"-3"$
$"E/mol·l"^"-1": color(white)(m)c"-1.00 × 10"^"-3"color(white)(mml)"1.00 × 10"^"-3"color(white)(mm)"1.00 × 10"^"-3"$

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

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

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

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

The $"HNO"_2$ will react completely with the $"NaOH"$ .

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

$color(white)(mmmmll)"HNO"_2 + "NaOH → NaX" + "H"_2"O"$
$"I/mol": color(white)(mll)0.35 color(white)(mmm)5.0$
$"C/mol": color(white)(m)"-0.35"color(white)(mml)"-0.35"$
$"E/mol": color(white)(mll)0color(white)(mmmm)4.6$

So, we have 4.6 mmol of $"NaOH"$ in 150 mL of solution.

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

$"pOH = -log"0.031 = 1.5$

$"pH = 14.00 - pOH = 14.00 - 1.5 = 12.5"$

All three solutions have the same pH.

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