Question

Problem 1,: 500 mL of a 0.500 M solution of NaOH is titrated with a 2 M acid solution. How much acid is needed to reach its titration end-point?

Problem 2: It takes 20.0mL of 4.00M NaOH to neutralize 5.00 mL of HCL solution. What is the concentration of the HCl solution?

Problem 3: 60. mL of a 1.0 M solution of H2SO4 has titrated to it endpoint with 1 L of a basic solution. What is the molarity of the base?

Problem 4: If I add acid to 100 mL of a 0.15 M NaOH solution until it is titrated with 150 mL of acid, what will the molarity of the acid solution be?

(Last Question I swear-)

Answer 1

Problem 1:

The balanced chemical equation for the reaction between NaOH and the acid is:

NaOH + HX → NaX + H2O

where X represents the acid.

The stoichiometry of the reaction shows that one mole of NaOH reacts with one mole of the acid. Thus, the number of moles of acid required can be calculated as follows:

Moles of NaOH = volume (in L) x concentration (in M)

Moles of NaOH = 500 mL x (1 L/1000 mL) x 0.500 M

Moles of NaOH = 0.250 moles

Since the number of moles of the acid required is the same as the number of moles of NaOH, the amount of acid needed to reach the end-point is:

Moles of acid = 0.250 moles

The volume of the acid required can be calculated using its concentration:

Moles of acid = volume (in L) x concentration (in M)

0.250 moles = volume (in L) x 2 M

Volume of acid = 0.125 L or 125 mL

Therefore, 125 mL of the 2 M acid solution is required to reach the titration end-point.

Problem 2:

The balanced chemical equation for the reaction between NaOH and HCl is:

NaOH + HCl → NaCl + H2O

The stoichiometry of the reaction shows that one mole of NaOH reacts with one mole of HCl. Thus, the number of moles of NaOH used in the titration can be calculated as:

Moles of NaOH = volume (in L) x concentration (in M)

Moles of NaOH = 20.0 mL x (1 L/1000 mL) x 4.00 M

Moles of NaOH = 0.080 moles

Since one mole of NaOH reacts with one mole of HCl, the number of moles of HCl present in the 5.00 mL sample can be calculated as:

Moles of HCl = Moles of NaOH

Moles of HCl = 0.080 moles

The concentration of the HCl solution can be calculated as follows:

Concentration of HCl = moles/volume (in L)

Concentration of HCl = 0.080 moles/(5.00 mL x 1 L/1000 mL)

Concentration of HCl = 16.0 M

Therefore, the concentration of the HCl solution is 16.0 M.

Problem 3:

The balanced chemical equation for the reaction between H2SO4 and the base is:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

The stoichiometry of the reaction shows that one mole of H2SO4 reacts with two moles of NaOH. Thus, the number of moles of NaOH used in the titration can be calculated as:

Moles of NaOH = volume (in L) x concentration (in M)

Moles of NaOH = 1.00 L x 1.0 M

Moles of NaOH = 1.0 moles

Since two moles of NaOH react with one mole of H2SO4, the number of moles of H2SO4 present in the solution is:

Moles of H2SO4 = 0.5 moles

The molarity of the base can be calculated as follows:

Molarity of base = moles/volume (in L)

Molarity of base = 0.5 moles/0.060 L

Molarity of base = 8.33 M