At room temperature (25°C), 45 mL of a 0.002 M acid solution reacts with 30 mL of a

Question

At room temperature (25°C), 45 mL of a 0.002 M acid solution reacts with 30 mL of a
NaOH (aq) solution (pH = 11.95), determine (through calculations) whether the acid
is phosphoric acid (H3PO4), acetic acid (CH3COOH), or carbonic acid (H2CO3).

DrBob222 · Accepted Answer

pH 11.95 means pOH = 2.05
pOH = -log(OH^-)
(OH^-) = 0.00891
0.00891 x 30 mL = about 0.27 (Note: Calculations give this as 0.267 but I've rounded to 0.27 because you're allowed only two significant figures.)
mmols acid = 45 x 0.002 = 0.09

CH3COOH + NaOH ==> CH3COONa + H2O
0.09 requires 0.09 but we used 0.27. Can't be CH3COOH.

H2CO3 + 2NaOH ==> Na2CO3 + 2H2O
0.09 requires 0.18 but we used 0.27. Can't be H2CO3.

H3PO4 + 3NaOH ==> Na3PO4 + 3H2O
0.09 requires 0.27 and that's what we used. Must be H3PO4.

Anonymous · Answer

Thank you so much! Not it makes sense!