Question

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"`?

Answer 1

All three solutions have the same pH.

Explanation:

`"HCl"` and `"HI"`

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

Let's call them `"HX"`.

If

`"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 `"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.