What is the pH of a buffer prepared by adding 0.506 mol of the weak acid HA to 0.305 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66*10^-7.pH = ____What is the \rm pH after 0.150 mol of \rm HCl is added to the buffer from Part A? Assume no volumechange on the addition of the acid.pH = _What is the \rm pH after 0.195 mol of \rm NaOH is added to the buffer from Part A? Assume novolume change on the addition of the base.pH = ________

A buffer's pH is easily calculated by using the Henderson-Hasselbalch equation which relates the pH of the buffer to the p K a of the weak acid in the buffer and the concentrations of the acid and its conjugate base in the buffer. It is mathematically expressed as:

p H = p K a + l o g[ A − ] / [ H A ]

First we will calculate the p K a of the weak acid from the given K a , as well as the concentration of HA and A − :

p K a = − log K a = − log ( 5.66 × 10 − 7 ) = 6.247 m o l e

H A =0.506m o l e HA / 2 L = 0.253M

[ A − ] = 0.305 m o l e A / 2.00 L = 0.153 M

Now we will use the Henderson-Hasselbalch equation to calculate the pH

pH = 6.247 + log ( 0.153 / 0.253)

= 6.247-0.21842909035

= 6.028

Part B

Because A − react with HCl in 1-to-1 molar ratio and HCl is the limiting reactant, all of the HCl added will react with completely with the A − in the buffer to form HA. The new concentration of HA and A − are:

[ H A ] = 0.150+0.506 / 2 = 0.328M

{A} = ( 0.305 − 0.150 ) m o l e A / 2.00 L = 0.0775 M

Applying the Henderson-Hasselbalch equation again

p H = 6.247 + log ( 0.0775 / o.328) = 5.620

PART C:

NaOH reacts with HA in 1-to-1 molar ratio and NaOH is the limiting reactant in this case, all the NaOH will react with some of the HA to form more

[ A − ] in the buffer. The new concentrations of HA and [ A − ] are:

[ H A ] = 0.506-0.195 / 2 = 0.1555 M

[A-] = 0.305 +0.195 / 2 = 0.250M

Applying the Henderson-Hasselbalch equation again

p H = 6.247 + log 0.250 / 0.1555

pH = 6.453

batolisis batolisis · Answer 2 · 2022-02-22T15:05:33+01:00

A) The pH of the buffer prepared is by adding 0.506 mol of the HA : 6

B ) The pH after 0.150 mol of HCl is added : 5.6

C ) The pH after 0.195 mol of NaOH is added : 6.5

A) Determine the pH of a buffer prepared by adding 0.506 mol of weak acid

Applying Henderson-Hasselbalch equation

pH = pKa + log[ A⁻ ] / [ HA ] ---- ( 1 )

First step : calculate the values of pKa , HA and A⁻

pKa = − logKa

= − log ( 5.66 × 10 − 7 )

= 6.247 mol

HA = 0.253M ( i.e. 0.506 / 2 )

A⁻ = 0.305 / 2.00 = 0.153 M

Insert values into equation ( 1 )

pH = 6.247 + log ( 0.153 / 0.253 )

= 6.028 ≈ 6

B) Determine the pH value after 0.150 mol of HCl is added

concentration of [ HA ] = ( 0.150 + 0.506 ) / 2

= 0.328 M

concentration of [ A ] = ( 0.305 - 0.150 ) / 2

= 0.0775 M

Insert values into equation ( 1 )

pH = 6.247 + log ( 0.0775 / 0.328 )

= 5.620 ≈ 5.6

C) Determine the pH value after 0.195 mol of NaOH is added

[ HA ] = ( 0.506 - 0.195 ) / 2

= 0.1555 M

[ A⁻ ] = ( 0.305 + 0.195 ) / 2

= 0.250M

Insert values into equation ( 1 )

pH = 6.453 ≈ 6.5

Hence we can conclude that A) The pH of the buffer prepared is by adding 0.506 mol of the HA : 6

B ) The pH after 0.150 mol of HCl is added : 5.6

C ) The pH after 0.195 mol of NaOH is added : 6.5

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