1 (a) Calculate the hydronium ion (H3O\') and hydroxide (OH) concentrations in a

Question

1 (a) Calculate the hydronium ion (H3O') and hydroxide (OH) concentrations in a solution prepared by diluting 0.100 mole of HCN to 1.00 liter with water. The acid dissociation constant (Ka) for HCN is 7.2.100 (b) Calculate the hydronium ion (H3O) and hydroxide (OH) concentrations in a solution prepared by diluting 0.100 mole of NaCN to 1.00 liter with water. NaCN is a strong electrolyte. (c) What is the pH of a solution prepared by adding 50.0 ml of 0.100 M NaOH to 200.0 ml of 0.200 M HCN? constant (K,) for HCN is 7.2 100

Explanation / Answer

a) HCN + H2O ---> H3O+ + CN-

0.1-x x x

Conc of HCN = 0.1/1 = 0.1 M

We just need to solve the dissociation equation = Ka = x^2 / 0.1-x

H+ comes out to be = square root (Ka*C )

= squareroot(7.2*10^-10 *0.1 )

= 8.48*10^-6 M

OH- = 10^-14 /  8.48*10^-6 = 1.17*10^-9 M

b) The solution shown above is correct for B

c) OH- = 0.05 * 0.1 = 0.005 moles

H+ = 0.2*0.2 = 0.04 moles

Therefore OH- will react with H+ to give water

Excess H+ = 0.04-0.005 = 0.035 moles

Conc = 0.035 / 0.250 = 0.14

pH = 0.85

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