Question

Question 3 

The solubility of Mg(OH)₂: (58.3 g/mol) is 9.63 mg Mg(OH)₂/100.0 mL solution at 25 °C. What is the poH of Mg(OH)₂ at this temperature?

Question 4 

What is the pH of a 0.7 mol/L benzoic acid (C₆H₅COOH, K_a =6.6x10^-5) solution?

Answer 1

solubility of base = 9.63 mg/100 ml =96.3 mg/ 1000 ml

= 96.3 mg of base in 1 l

96.3 mg of Mg(OH)2 = 96.3*10^-3/58

= 1.66*10^-3

Mg(OH)2 -> Mg2+ + 2 (OH)-

1 mol Mg(OH)2 produces 2 moles of (OH)-

moles of (OH)- = 2 *1.96 * 10^-3

=.00392 in 1 l sol

$pOH=-log(.00392)$

$pOH=2.4$

For weak acid following formula can be uswd direcly for calculating pH

pKa= -log(Ka)

= -log(6.6*10^-5)

= 4.18

pH= 2.167

Answer 2

solubility of base = 9.63 mg/100 ml =96.3 mg/ 1000 ml

= 96.3 mg of base in 1 l

96.3 mg of Mg(OH)2 = 96.3*10^-3/58

= 1.66*10^-3

Mg(OH)2 -> Mg2+ + 2 (OH)-

1 mol Mg(OH)2 produces 2 moles of (OH)-

moles of (OH)- = 2 *1.96 * 10^-3

=.00392 in 1 l sol

$pOH=-log(.00392)$

$pOH=2.4$

For weak acid following formula can be uswd direcly for calculating pH

pKa= -log(Ka)

= -log(6.6*10^-5)

= 4.18

pH= 2.167