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

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Answer 1 · 2018-04-02T13:46:29

#"0.102 g"#

#"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 · 2018-04-02T14:21:11

Around #102# milligrams.

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"#