Question

In: Chemistry

I got the answer for part A right but I can't seem to get part B...

I got the answer for part A right but I can't seem to get part B or C. Could you please show me working out? Thank you!

Part A

What is the pH of a buffer prepared by adding 0.506mol of the weak acid HA to 0.609mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66

Expert Solution

Part B

Addition of HCl will convert NaA into HA

mol of HCl = 0.150 mol

mol of HA = 0.506 mol

mol of NaA = 0.609 mol

pKa of HA = -log( 5.66 x 10^-7) = 6.247

NaA + HCl <--------> HA + NaCl

0.609 0.150 0.506 ( Initial)

- 0.150 - 0.150 + 0.150 (Change)

0.459 0 0.656 (Final)

pH = pKa + log([NaA] / [HA])

pH = 6.247 + log(0.459 / 0.656)

pH = 6.247 - 0.155

pH = 6.092

Part C

NaOH reacts with HA

mol of NaOH = 0.195 mol

mol of HA = 0.506 mol

mol of NaA = 0.609 mol

HA + NaOH ----------> NaA + H2O

0.506 0.195 mol 0.609 mol (Initial)

- 0.195 - 0.195 + 0.195 (Change)

0.311 0 0.804

pH = pKa + log([NaA] / [HA])

pH = 6.247 + log(0.804 / 0.311)

pH = 6.247 + 0.412

pH = 6.659

queen_honey_blossom answered 3 hours ago

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