You have a 0.125 M solution of benzoic acid. You now add enough sodium benzoate

Question

You have a 0.125 M solution of benzoic acid. You now add enough sodium benzoate to get to a benzoate concentration of 0.15 M. What is the pH of the resulting buffer solution You have a solution that contains 0.10 M acetic acid and 0.15 M acetate. What is the pH of this buffer You add 0.02 moles of OH to 1 L of this buffer. Recalculate the pH. You add 0.03 moles of H^+ to 1L of the original buffer. Recalculate the pH. Calculate the pH of: (5 points each) a a buffer containing 0.12 M HCIO and 0.1 M NaCIO. recalculate the pH after adding 25 mL of 0.1 M HCI to 250 mL of the buffer in a). recalculate the pH after adding 15 mL of 0.1 M NaOH to 250 mL of the buffer in a). What is the pH of a solution prepared by adding 100 mL of a 0.1 M NaOH solution to 250 mL of a 0.1 M HCN solution How many grams of K_2HPO_4 must be added to 2 L of a 0.1 M KH_2PO_4 solution to produce a pH = 7.0 buffer Identify the acid and the conjugate base. How many grams of NaOH must be added to 1 L of a 0.1 M KH_2PO_4 to prepare a pH 7.42 buffer. Calculate the pH of a buffer prepared by adding 8.0 g of ammonium nitrate. NH_4NO_3, to 1 L of a 0.12 M solution of ammonia, NH_3. (This is a buffer made by adding the conjugate acid, NH_4^+, to a weak base, NH_3.)

Explanation / Answer

(2):

pKa of benzoic acid(C6H5COOH) = 4.202

[C6H5COOH] = 0.125 M

[C6H5COO-] = 0.15 M

Applying Hendersen equation

pH = pKa + log[C6H5COO-] / [C6H5COOH] = 4.202 + log(0.15 M / 0.125 M) = 4.28

=> pH = 4.28 (answer)

(3): (a): pKa of acetic acid(CH3COOH) = 4.76

[CH3COOH] = 0.10 M

[CH3COO-] = 0.15 M

Applying Hendersen equation

pH = pKa + log[CH3COO-] / [CH3COOH] = 4.76 + log(0.15 M / 0.10 M) = 4.92

=> pH = 4.92 (answer)

(b): For 1L of the buffer, moles of CH3COOH = 0.10 mol

moles of CH3COO- = 0.15 mol

When we add 0.02 mol of OH- ion it will react with 0.02 mol of CH3COOH to form 0.02 mol of CH3COO-.

-------------CH3COOH + OH- ------- > CH3COO- + H2O

Init.mol: 0.10, ----------- 0.02 ----------- 0.15

chaange: - 0.02, ------- - 0.02, -------- + 0.02

eqm.mol: 0.08, --------- 0 -------------- 0.17

eqm.conc: 0.08 M, -----0 M ----------- 0.17 M

Applying Hendersen equation

pH = pKa + log[CH3COO-] / [CH3COOH] = 4.76 + log(0.17 M / 0.08 M) = 5.07

=> pH = 5.07 (answer)

(c):

When we add 0.03 mol of H+ ion it will react with 0.03 mol of CH3COO- to form 0.03 mol of CH3COOH.

-------------CH3COO- + H+ ------- > CH3COOH

Init.mol: 0.15, ---------0.03 ----------- 0.10

chaange: - 0.03, --- - 0.03, -------- + 0.03

eqm.mol: 0.12, --------- 0 -------------- 0.13

eqm.conc: 0.12 M, -----0 M ----------- 0.13 M

Applying Hendersen equation

pH = pKa + log[CH3COO-] / [CH3COOH] = 4.76 + log(0.12 M / 0.13 M) = 4.71

=> pH = 4.71 (answer)

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