Question

A.) Find the pH of a solution that contains 3.25 g of H2SO4 dissolved in 2.75 liters of solution.

B.) If the pH is 9.85, what is the concentration of the aluminum hydroxide solution?

Answer 1

Part (A)

Data given : Weight of H₂SO₄ = 3.25 grams

                 Volume of solution = 2.75 litres

Calculation of molarity of H₂SO₄ solution

Molar mass of H₂SO₄ = 98 grams/moles

Number of moles of H₂SO₄ = weight of H₂SO₄ / molar mass of H₂SO₄

                                        = 3.25/98 = 0.03316326 moles

Molarity (M) = moles oF H₂SO₄/ Volume of solution (in litre)

                  = 0.03316326 / 2.75 = 0.01206 M

Since sulfuric acid is a strong acid the concentration of H⁺ ion produced will be same as concentration of H₂SO₄

H₂SO₄       --------> 2H⁺                 +         SO₄^2-

0.01206                  2( 0.01206)

                             =0.02412

since 1 moles of H₂SO₄ produces 2 moles of H⁺ ion

Concentration of H⁺ = 2* concentration of H₂SO₄ = 2*0.01206 =0.02412 M

[H⁺] = 0.02412 M

Now pH is calculated with the help of following relation

pH = - Log([H+] ) = - Log(0.02412) = 1.618

Hence pH of solution that contains 3.25 g of H₂SO₄ dissolved in 2.75 litres of solution is 1.618

Part (B)

Data given: pH =9.85

Now let us calculate pOH from this data,

We know that,

pH + pOH =14

pOH = 14 - pH = 14-9.85 = 4.15

Let us now convert pOH value in to molar concentration

pOH = - Log([OH^-])

4.15 = - Log([OH^-])

[OH^-] = 10^-4.15 =7.079*10^-5 M

Now let us write the dissociation equation for the given compound

Al(OH)₃ ------> Al³⁺ + 3 OH ^-

1 moles of Al(OH)₃ = 3 moles of OH^-

Hence concentration of Al(OH)₃ = (1/3) * concentration of OH^- = (1/3)*7.079*10^-5 = 2.36 * 10^-5 M

Hence, when pH =9.85, Concentration of aluminum hydroxide solution is 2.36 * 10^-5 M