If you titrate 12.6 mL of an unknown monoprotic (one H) acid with 13.0 mL of 0.140 M NaOH, what will be the concentration of the acid?
If you add 23.5g of NaCl to 1.2 kg of water, what will be the boiling point and the freezing point of the new solution? (Kf=1.86, Kb=0.52)
((i appreciate every help i can get, tho i prefer to know how to solve for it)) THANK you so much
<If you titrate 12.6 mL of an unknown monoprotic (one H) acid with 13.0 mL of 0.140 M NaOH, what will be the concentration of the acid?>
As the missing information is the concentration of the unknown acid I assume that is what is being asked for here. The question is poorly worded.
As always start with a balanced equation
HA + NaOH -> NaA + H2O
so 1 mole of HA reacts with 1 mole of NaOH
number of millimoles of NaOH used =
13.0 x 0.140
which must be the same as the original number of millimoles fo HA in solution = 12.6 x X
13.0 x 0.140 = 12.6 x X
find X the concentration of the acid