A.) A buffer solution contains 0.376 M Na HCO 3 and 0.386 M Na 2 CO 3 . If 0.017

Question

A.) A buffer solution contains 0.376 MNaHCO3 and 0.386 M Na2CO3.

If 0.0175 moles of sodium hydroxideare added to 125 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding sodium hydroxide)

pH =

B.) A buffer solution contains 0.329 Mnitrous acid and 0.437 M potassiumnitrite.

If 0.0320 moles of hydrobromic acidare added to 250. mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume change does not change upon adding hydrobromic acid)

pH =

C.) A buffer solution contains 0.271 Mammonium bromide and 0.306 Mammonia.

If 0.0477 moles of perchloric acid are added to 225 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume change does not change upon adding perchloric acid)

pH =

Explanation / Answer

A.)

A buffer solution contains 0.376 M NaHCO3 and 0.386 M Na2CO3.

Pka2 of H2CO3 i.e. pKa =of HCO3– = 10.32

On addition 0.0175 moles of sodium hydroxide to 125 mL = 0.125L of this buffer,

[Na2CO3] will increase and [NaHCO3] will decrease.

[Na2CO3] = (0.386 x 125mL + 0.0175 x 125 mL) milimoles = (0.386+0.0175) x 125 milimoles. = 0.4035 x 125 milimoles

[NaHCO3] = (0.376 x 125mL – 0.0175 x 125 mL) milimoles = (0.376 – 0.0175) x 125 milimoles = 0.3585 x 125 milimoles.

Using Henderson – Hasselbalch equation pH will be given as,

pH = pKa + log ([salt]/[Acid])

pH = pKa of HCO3– + log ([Na2CO3]/[ HCO3-])

pH = 10.32 + log {(0.4035 x 125)/(0.3585 x 125)}

pH = 10.32 + log(1.265)

pH = 10.32 + 0.10

pH = 10.42

pH of solution on adding

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B)For HNO2 – NaNO2 buffer solution

[HNO2] =0.329 M and [NaNO2] = 0.437 M

On adding 0.0320 moles of HCl to 250 mL of this buffer there will be increase in [HNO2] and decrease in [NaNO2] as hCl is strong acid and ionizes H+ ions which protonated NO2– ions back to HNO2. Hence new concentrations are

[HNO2] = (0.329 x 250 + 0.0320 x 250 mL) milimoles = 0.361 x 250 milimoles

[NaNO2] = 0.437 x 250 – 0.0320 x 250 mL) milimoles = 0.405 x 250 milimoles.

According to Henderson – Hasselbalch equation,

pH = pKa of HNO2 + log ([NaNO2]/[HNO2])

pH = 3.39 + log{(0.405 x 250)/(0.361 x 250)}

pH = 3.39 + log(0.405/0.361)

pH = 3.39 + log(1.1219)

pH = 3.39 + 0.05

pH = 3.44

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C.) For NH4Br – NH3 buffer solution

[NH4Br] = [NH4+]= 0.271 M NH4Br

[NH3] = 0.306 M NH3.

On adding 0.0477 moles of perchloric acid are added to 225 mL of this buffer, [NH4+] will increase and of [NH3] will decrease.

[NH4+] = (0.271 + 0.0477 ) x 225 = 0.3187 x 225 milimoles

[NH3] = (0.306 – 0.0477) x 225 = 0.2583 x 225 milimoles.

By Henderson- Hasselbalch equation,

pH = pKa of NH4+ + log([NH3]/[NH4+]) ……………(NH4+ is acid)

pH = 9.26 + log {(0.2583 x 225)/(0.3187 x 225)}

pH = 9.26 + log (0.2583/0.3187)

pH = 9.26 + log(0.8105)

pH = 9.26 + (– 0.09)

pH = 9.17

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