Question

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 = ______________

Answer 1

To determine the pH of a buffer solution, use the Henderson-Hasselbalch equation. For the original buffer, the pH is 6.03. When 0.150 mol HCl or 0.195 mol NaOH is added, the concentrations of HA and A- change, and the pH can be recalculated using the new concentrations in the Henderson-Hasselbalch equation.

Step-by-step explanation:

To calculate the pH of the buffer solution, we will use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). Given that the concentration of the weak acid HA is 0.506 mol/2.00 L = 0.253 M, and the concentration of its conjugate base NaA is 0.305 mol/2.00 L = 0.1525 M, and the dissociation constant Ka of HA is 5.66*10^-7, we first calculate the pKa by taking the negative logarithm of Ka: pKa = -log(Ka).

Therefore, pKa = -log(5.66*10^-7) = 6.25. Substituting the concentrations and the pKa into the Henderson-Hasselbalch equation: pH = 6.25 + log(0.1525/0.253), which gives pH = 6.25 + log(0.6024) = 6.25 - 0.2199 = 6.03.

When 0.150 mol of HCl is added to the buffer, it reacts with the conjugate base NaA. The new concentrations will be [HA] = 0.506 mol + 0.150 mol = 0.656 mol / 2 L = 0.328 M and [A-] = 0.155 mol / 2 L = 0.0775 M. Again, using the Henderson-Hasselbalch equation, we can find the new pH.

For the addition of 0.195 mol of NaOH to the buffer, NaOH will react with the HA. The new concentrations after the reaction will be [HA] = 0.506 mol - 0.195 mol = 0.311 mol / 2 L = 0.1555 M and [A-] = 0.305 mol + 0.195 mol = 0.500 mol / 2 L = 0.250 M. Using the Henderson-Hasselbalch equation again, we can find the pH after this addition.

Answer 2

A. pH= 6.028

b. pH = 5.620

C .pH = 6.453

Step-by-step explanation:

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