A 20.0 mL sample of 0.432 M HBr is titrated with a 0.314 M NaOH
solution. The pH after the addition of 35.0 mL of NaOH is
__________.
A.0.04
B.1.37
C.12.53
D.12.63
E.13.33
A 35.00 mL sample of an unknown H3PO4solution is titrated with a
0.130 M NaOH solution. The equivalence point is reached when 25.83
mLof NaOH solution is added.
What is the concentration of the unknown H3PO4 solution? The
neutralization reaction is
H3PO4(aq)+3NaOH(aq)→3H2O(l)+Na3PO4(aq)
a 22.5 ml sample of an acetic acid solution is titrated with a
0.175 M NaOH solution. The equivalence point is reached when 37.5
ml of the base is added. What was the concentration of acetic acid
in the original 22.5 ml? what is the ph of the equivalence point?
(ka (acetic acid) = 1.75 x 10^-5)
A25.0 mL sample of 0.100 M NaOH is titrated with a 0.100 M HNO3
solution. Calculate the pH after the addition of 0.0, 4.0, 8.0,
12.5, 20.0, 24.0, 24.5, 24.9, 25.0, 25.1, 26.0, 28.0, and 30.0 of
the HNO3.
50.0 mL of a 0.100 -M HoAc solution is titrated with a 0.100 -M
NaOH solution. Calculate the pH at each of the following
points.
Volume of NaOH added: 0, 5, 10, 25, 40 ,45 , 50 , 55 , 60 , 70 ,
80 , 90 , 100
a 2.00 mL sample of vinegar is titrated with 15.86 mL of 0.350
M NaOH to a phenolphthalein end point
A) calculate the molar its of acetic acid in the vinegar
solution
B) calculate the % of acetic acid in the vinegar
A 15.3 mL sample of vinegar, containing acetic acid, was
titrated using 0.809 M NaOH solution. The titration required 24.06
mL of the base. Assuming the density of the vinegar is 1.01 g/mL,
what was the percent (by mass) of acetic acid in the vinegar?
A 12.9 mL solution of 0.100 mol L-1 HOCl is titrated
using 0.150 mol L-1 NaOH.
What is the pH of the solution after 5.18 mL of the NaOH
solution is added? Express your answer to 2 decimal places.
You have 5 attempts at this question.
Remember you can find KA and/or KB values
in your textbook in chapter 15.