Question

The reaction of HCl with NaOH is represented by the equation

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

What volume of 0.555 M HCl is required to titrate 17.7 mL of 0.359 M NaOH?

a) 3.53 mL b) 1.80mL c) 11.4mL d) 17.7mL e)27.4 mL

In a volumetric analysis experiment, a solution of sodium oxalate (Na₂C₂O₄) in acidic solution is titrated with a solution of potassium permanganate (KMnO₄) according to the following balanced chemical equation:

2KMnO₄ (aq) + 8H₂SO₄ (aq) + 5Na₂C₂O₄ (aq) → 2MnSO₄ (aq) +8H₂O (l) +10CO₂ (g) + 5Na₂SO₄ (aq) + K₂SO₄ (aq)

What volume of 0.0388 M KMnO₄ is required to titrate 0.134 g of Na₂C₂O₄ dissolved in 20.0 mL of solution?

a) 1.38 mL b) 3.45mL c) 10.3 mL d) 25.8 mL e) 20.0 mL

In order to dilute 35.5 mL of 0.533 M HCl to 0.100 M, the volume of water that must be added is

a) 28.8 mL. b) 6.66 mL c) 189mL d) 0.00150 mL e) 154mL

The concentration of sulfate in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of the wastewater is added an excess of calcium nitrate, forming the insoluble calcium sulfate (MM = 136.1 g/mol) according to the balanced equation given below. The solid calcium sulfate is dried, and its mass is measured to be 0.7485 g. What was the concentration of sulfate in the original wastewater sample?

SO₄^2-(aq) + Ca(NO₃)₂(aq) → CaSO₄(s) + 2NO₃^-(aq)

a) 1.818 M b) 10.19 M c) 0.05500 M d) 18.8 M

Answer 1

The reaction of HCl with NaOH is represented by the equation

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

What volume of 0.555 M HCl is required to titrate 17.7 mL of 0.359 M NaOH?

Solution :-

Mole ratio of the NaOH and HCl is 1 : 1

Therefore

Volume of HCl = volume of NaOH * molarity of NaOH / molarity of HCl

                         = 17.7 ml * 0.359 M / 0.555 M

                         = 11.4 ml

So correct answer is option C that is 11.4 ml

a) 3.53 mL b) 1.80mL c) 11.4mL d) 17.7mL e)27.4 mL

In a volumetric analysis experiment, a solution of sodium oxalate (Na₂C₂O₄) in acidic solution is titrated with a solution of potassium permanganate (KMnO₄) according to the following balanced chemical equation:

2KMnO₄ (aq) + 8H₂SO₄ (aq) + 5Na₂C₂O₄ (aq) → 2MnSO₄ (aq) +8H₂O (l) +10CO₂ (g) + 5Na₂SO₄ (aq) + K₂SO₄ (aq)

What volume of 0.0388 M KMnO₄ is required to titrate 0.134 g of Na₂C₂O₄ dissolved in 20.0 mL of solution?

Solution :-

Lets first calculate the moles of the Na2C2O4

Moles = mass / molar mass

Moles of Na2C2O4 = 0.134 g / 134 g per mol

                          = 0.001 mol

Now using the mole ratio lets calculate the moles KMnO4

0.001mol Na2C2O4 * 2 mol KMnO4 / 5 mol Na2C2O4 = 0.0004 mol

Now lets calculate the volume of the KMnO4

Volume = moles / molarity

              = 0.0004 mol / 0.0388 mol per L

             = 0.0103 L

0.0103 L * 1000 ml / 1 L = 10.3 ml

So the answer is option ‘C’ that is 10.3 ml

a) 1.38 mL b) 3.45mL c) 10.3 mL d) 25.8 mL e) 20.0 mL

In order to dilute 35.5 mL of 0.533 M HCl to 0.100 M, the volume of water that must be added is

Solution:-

Lets calculate the final volume of the dilute solution

M1V1 = M2V2

V2 = M1V1/M2

      = 0.533 M * 35.5 ml / 0.100 M

     = 189.2 ml

So the volume of water needed = 198.2 ml – 35.5 ml = 153.7 ml = 154 ml

So the correct answer is option ‘e’ that is 154 ml water is needed

a) 28.8 mL. b) 6.66 mL c) 189mL d) 0.00150 mL e) 154mL

The concentration of sulfate in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of the wastewater is added an excess of calcium nitrate, forming the insoluble calcium sulfate (MM = 136.1 g/mol) according to the balanced equation given below. The solid calcium sulfate is dried, and its mass is measured to be 0.7485 g. What was the concentration of sulfate in the original wastewater sample?

SO₄^2-(aq) + Ca(NO₃)₂(aq) → CaSO₄(s) + 2NO₃^-(aq)

Solution :- lets calculate the moles of calcium sulfate

Moles of CaSO4 = 0.7485 g / 136.1 g per mol = 0.0055 mol

Mole ratio of the CaSO4 to SO4^2- is 1 : 1

So the moles of sulfate reacted = 0.0055 mol

Now lets calculate the molarity of the sulfate

Molarity of SO4^2- = 0.0055 mol / 0.100 L

                               = 0.05500 M

So the correct answer is option ‘C’ that is 0.05500 M

a) 1.818 M b) 10.19 M c) 0.05500 M d) 18.8 M