A buffer is prepared by adding 0.506mol of the weak acid HA to 0.609 mol of NaA

Question

A buffer is prepared by adding 0.506mol of the weak acid HA to 0.609 mol of NaA in 2.00L of solution. The dissociation constant Ka of HA is 5.66 x 10 ^ -7. pH = 6.33

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

Explanation / Answer

[HA] = 0.708/ 2.00 = 0.354 M [A-] = 0.609/2.00 =0.305 M pKa = 6.25 pH = 6.25 + log 0.305 / 0.354 =6.19 A- + H+ >> HA moles A- = 0.609 - 0.150 =0.459 [A-]= 0.459/ 2.00 =0.230 M moles HA = 0.708 + 0.150 =0.858 [HA] = 0.858/2.00 = 0.429 M pH = 6.25 + log 0.230/0.429 = 5.98 HA + OH- >> A- + H2O moles HA = 0.708 - 0.195 =0.513 [HA]= 0.513/2.00 =0.257 M moles A- = 0.609 + 0.195 =0.804 [A-]= 0.804/2.00 = 0.402 M pH = 6.25 + log 0.402/0.257 =6.44

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