24.56 mL of a H2SO4 solution of unknown molarity is titrated to the equivalence point with...

24.56 mL of a H2SO4 solution of unknown molarity is titrated to the equivalence point with 19.3mL of a 0.45 M KOH solution to the equivalence point. What is the concentration of the acid solution? (please show work and answer in correct sig figs and scientific notation)

Expert Solution

H2So4 (aq) + 2KOH (aq) -------------------> K2So4(aq) + 2H2O(l)

1 mole 2 moles

H2So4 KOH

M1 = M2 = 0.45M

V1 = 24.56ml V2 = 19.3ml

n1 = 1 n2 = 2

M1V1/n1 = M2V2/n2

M1 = M2V2n1/V1n2

= 0.45*19.3*1/24.56*2 = 0.18 M

The conc of H2So4 is 0.18M = 1.8*10^-1M

queen_honey_blossom answered 1 year ago

A solution of HF is titrated with 0.150 M NaOH. The pH at the equivalence point...

A solution of HF is titrated with 0.150 M NaOH. The pH at the equivalence point will be greater than 7.0 less than 7.0 equal to 7.0 cannot be estimated from the information given.

A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence...

A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence point is reached after 21.27 mL of the base were added. Calculate 1) the concentration of the acid in the original solution, 2) the pH of the original HCl solution and the original NaOH solution, 3) the pH after 10.00 mL of NaOH have been added, 4) the pH at the equivalence point, and 5) the pH after 25.00 mL of NaOH have been...

Determine the pH at the equivalence point when 0.05 M KOH is titrated with 50.0 ml...

Determine the pH at the equivalence point when 0.05 M KOH is titrated with 50.0 ml of 0.01 M benzoic acid HC7H5O2. The Ka for benzoic acid is 6.3 x10-5.

0.1 M NaOH is titrated with 25 mL acetic acid. The equivalence point was reached at...

0.1 M NaOH is titrated with 25 mL acetic acid. The equivalence point was reached at 23.25 mL. a) Based on your experimentally observed equivalence point of 23.25 mL, calculate the original concentration of the acetic acid that came from the stock bottle. b) Based on your experimentally observed half equivalence point of 11.625 mL pH= 4.8, calculate the Ka of acetic acid. c) Based on your answers to 1& 2 calculate the expected pH at the equivalence point

A 30.00 mL solution of 0.235 M CH2NH2 was titrated with 0.150 M H2SO4. a) Calculate...

A 30.00 mL solution of 0.235 M CH2NH2 was titrated with 0.150 M H2SO4. a) Calculate the pH of the 30.00 mL solution of 0.235 M CH2NH2 b) Calculate the volume of 0.150 M H2SO4 required to neutralize the 30 mL solution of 0.235 M CH2NH2. c) Calculate the volume of 0.150 M H2SO4 required to neutralize half the mole of CH2NH2 present 30 Ml solution of 0.235 M CH2NH2 d) Calculte the pH of the solution in (c). e)...

what is the molarity of a 50ppm (wt/vol) solution of H2SO4?

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)

0.901g KHP is fully titrated with 44.1 mL of an NaOH solution of unknown concentration. What...

0.901g KHP is fully titrated with 44.1 mL of an NaOH solution of unknown concentration. What is the concentration of the NaOH solution? (Report to 3 sig figs).

96.0 mL of a NaOH solution of unknown concentration is titrated with 4.00 M hydrochloric acid,...

96.0 mL of a NaOH solution of unknown concentration is titrated with 4.00 M hydrochloric acid, HCl. The end point is reached when 150.0 mL of acid are added to the base. What is the concentration of the original NaOH solution? a. 1.56 mol/L b. 3.13 mol/L c. 6.26 mol/L d. 12.5 mol/L

What is the pH at the equivalence point when 95.0 mL of a 0.275 M solution...

What is the pH at the equivalence point when 95.0 mL of a 0.275 M solution of acetic acid (CH3COOH) is titrated with 0.100 M NaOH to its end point?

Question