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Part A What is the pH of a buffer prepared by adding 0.809 mol of the...

Part A What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.608 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7. Part B What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. (Express answer to 3 decimal places) Part C What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

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Expert Solution

Part A What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.608 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10?7.

Molarity of acid and salt are as follows:

Acid =0.809 /2.0 = 0.405 M

Salt = 0.608 /2.0 = 0.304 M

Ka of HA is 5.66×10?7

pKa = 6.25

pH = pKa + log(A-/HA)

pH =6.25 + log(0.304/0.405)

pH =6.25 + log 0.751

pH =6.25 -0.125

pH =6.125

Part B What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. (Express answer to 3 decimal places)

after addition of 0.150 mol HCl

A- decreases from 0.608 mol to 0.458 mol per 2L, so new molarity = 0.229 M

the HA increases from 0.607 mol to 0.757 mol , so new molarity =0.379 M

pH = pKa + log(A-/HA)

pH =6.25 + log(0.229/0.379)

pH =6.25 + log 0.604

pH =6.25 -0.219

pH =6.031

Part C What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

A- increases from 0.608 mol to 0.803 mol; , so new molarity =0.803/2.0=0.402 M

HA decreases from 0.607 mol to 0.412 mol; so new molarity =0.412/2.0=0.206 M

pH = pKa + log(A-/HA)

pH =6.25 + log(0.402/0.206)

pH =6.25 + log 1.95

pH =6.25 +0.290

pH =6.54

queen_honey_blossom answered 3 weeks ago
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