a) What is the pH of a buffer prepared by adding 0.607 mol of the weak acid HA t

Question

a) What is the pH of a buffer prepared by adding 0.607 mol of the weak acid HA to 0.507 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×107.

Express the pH numerically to three decimal places.

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 the pH numerically to three decimal places.

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.

Express the pH numerically to three decimal places.

Explanation / Answer

PH= PKa + log [salt]/[acid] = PH= PKa + log [NaA]/[HA]

The dissociation constant Ka of HA is 5.66×107

PKa = - log 5.66 x 10-7=6.25

[NaA ]= 0.507mol / 2.00 L= 0.254 M

[HA]= 0.607 mol / 2.00 L =0.303 M

pH = 6.25 + log 0.254 / 0.303= 6.17

NaOH + HCl= NaCl +H2O ,

Hence [NaOH] decreases hence [NaA] decreases[ and [HA] increases

HCl added to buffer = 0.150 mol

moles of [NaA] = 0.507- 0.150=0.357

[NaA]= 0.357/2.00 = 0.179 M

moles HA= 0.607 + 0.150=0.757

[HA]= 0.757 / 2.00=0.378 M

pH = 6.25 + log 0.179/ 0.378= 5.93

NaOH + HA= NaA +H2O Hence [HA] decreases hence [NaA] increases

moles HA = 0.607 - 0.195 = 0.412

[HA]= 0.412/2.00 = 0.206 M

[NaA ]= 0.507 + 0.195=0.702

[NaA]= 0.702/2.00=0.351 M

pH = 6.25 + log 0.351/ 0.206 = 6.48

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