Map We are going to work the calculation of the pH of a buffer prepared by mixin

Question

Map We are going to work the calculation of the pH of a buffer prepared by mixing a strong base with a weak acid. We want you to see all the steps of this type of problem by just answering the questions in stages. Make sure you can do the moles, then check your answer. Then do the where you need practice. This is a good pull-it-all-together type of problem. Enjoy! lem in one place. You can learn the pieces of this check your answer. Think about all the pieces and where you are strong and Compute the pH of the solution after the reaction of 42mL of 0.68M HCN with 27mL of 0.39 KOH HCN(aq) + KOH(aq) HOH(1) + KCN(aq) Compute the moles of reactants before reaction Number Number Moles of acid Moles of base moles Moles of reactant and pH active product after reaction Number Number Excess reactant Product moles moles Now convert moles to concentration units Number Number Reactant] Molar [Product] Molar

Explanation / Answer

Moles of HCN = 0.042 L * 0.68 M

= 0.02856 moles

Moels of KOH= 0.027 L* 0.39M= 0.01053 moles

Moles of acid present in excess = 0.02856-0.01053= 0.01803 Mol HCN

Moles of product KCN=

0.01053 moles KOH *1 mole KCN/ 1 mole KOH= 0.01053 KCN

Molar concentration after reaction = 0.01803 Mol HCN / total volume( 0.042 L +0.027 L

= 0.26 M

Molar concentration of product = 0.01053 Mol KCN / total volume( 0.042 L +0.027 L

= 0.15 M

Contrition of HCN = 0.01803 Mol HCN / total volume( 0.042 L +0.027 L

= 0.26 M

HCN < = > H+ CN-

Ka = [H+]² / [HCN]
3.5*10^-4 = [H+]² / 0.26
[H+]² = (3.5*10^-4) *0.26
[H+]² = 9.1*10^-5
[H+] = 9.54*10^- 3

pH = -log [H+]
pH = -log (9.54*10^- 3)
pH = 2.02

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---