Question

In: Chemistry

A 40 ml sample of .50 M solution of HOCl is titrated with .20 M NaOH...

A 40 ml sample of .50 M solution of HOCl is titrated with .20 M NaOH

a) what is the initial pH of the acidic solution?
b) what volume of base is necessary to reach the half equivalence point?
c) what is the pH at the half equivalence point?
d) what volume of base is necessary to reach the equivalence point?

Solutions

Expert Solution


Related Solutions

A 20.0 mL sample of 0.432 M HBr is titrated with a 0.314 M NaOH solution....
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....
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...
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...
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....
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 20.00 mL sample of 0.120 M NaOH was titrated with 52.00 mL of 0.175 M...
A 20.00 mL sample of 0.120 M NaOH was titrated with 52.00 mL of 0.175 M HNO3.             Calculate the [H+], [OH-], pH, and pOH for the resulting solution.   
A 50.0 -mL sample of 0.50-M acetic acid, CH3COOH, is titrated with a 0.150M NaOH solution....
A 50.0 -mL sample of 0.50-M acetic acid, CH3COOH, is titrated with a 0.150M NaOH solution. Calculate the pH at equivalence point. (Ka=1.8x10^-5)
a 2.00 mL sample of vinegar is titrated with 15.86 mL of 0.350 M NaOH to...
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 25.0 mL solution if 0.10 M acetic acid is titrated with 0.1 M NaOH solution....
a 25.0 mL solution if 0.10 M acetic acid is titrated with 0.1 M NaOH solution. the pH equivalence is
A 15.3 mL sample of vinegar, containing acetic acid, was titrated using 0.809 M NaOH solution....
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?
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT