Question

How many grams of `HCL` must you dissolve in `1400\ mL` to make a solution with `pH=2.7``\ \`?

Answer 1

`"0.102 g"`

Explanation:

`"pH" = -log["H"^+]`

`2.7 = -log["H"^+]`

`["H"^+] = 10^-2.7 ≈ "0.002 M"`

`"Molarity" = "Moles of solute"/"Volume of solution (in litres)"`

`"0.002 M" = "n"/"1.4 L"`

`"n = 0.0028 mol"`

Mass of `"HCl" = 0.0028 cancel"mol" × "36.46 g"/cancel"mol" = "0.102 g"`

Answer 2

Around `102` milligrams.

Explanation:

Well, we first address the `"pH"` equation, which states that,

`"pH"=-log[H^+]`

• `[H^+]` is the hydrogen ion concentration in terms of molarity

And so,

`2.7=-log[H^+]`

`[H^+]=10^-2.7`

`~~2*10^-3 \ "M"`

Therefore, the hydrochloric acid solution has a concentration of `2*10^-3 \ "M"=2*10^-3 \ "mol/L"`.

We want to make a `1400 \ "mL"=1.4 \ "L"` solution, so we must use:

`(2*10^-3 \ "mol")/(color(red)cancelcolor(black)"L")*1.4color(red)cancelcolor(black)"L"=2.8*10^-3 \ "mol"`

Hydrochloric acid has a molar mass of `36.46 \ "g/mol"`. So here, we need,

`2.8*10^-3color(red)cancelcolor(black)"mol"*(36.46 \ "g")/(color(red)cancelcolor(black)"mol")=0.102088 \ "g"`

`~~102 \ "mg"`