Respuesta :
Answer:
0.025 mol
Explanation:
When the NaH₂PO₄ reacts with NaOH, first it will dissociate in Na⁺ and H₂PO₄⁻, in a 1:1:1 reaction
NaH₂PO₄ → Na⁺ + H₂PO₄⁻
The number of moles of H₂PO₄⁻ is the number of moles of NaH₂PO₄
n = 0.250 L * 1 mol/L = 0.250 mol
The number of moles of NaOH that reacts is:
n = 0.150 L * 1 mol/L = 0.150 mol
The base will react with the acid H₂PO₄⁻ and will form the anion HPO₄²⁻ which is the conjugate base of the acid. The acid and it's conjugate base will form the buffer.
Let's suppose that all the base reacts, so, the number of moles of the substances of interest will be:
nH₂PO₄⁻ = 0.250 - 0.150 = 0.100 mol
nHPO₄²⁻ = 0.150 mol
The pH of a buffer can be calculated by the Handerson-Halsebach equation:
pH = pKa + log[A⁻]/[HA]
Where pKa = -logKa, [A⁻] is the conjugate base concentration (HPO₄²), and [HA] is the acid concentration (H₂PO₄⁻)
If the pH change for 0.18 units, the concentrations of A⁻ and HA will change too. So, let's call them [A⁻]f and [HA]f. Because an acid is being added, the pH must decrease, so pH final - pH initial = -0.18
(pKa + log[A⁻]f/[HA]f) - (pKa + log[A⁻]/[HA]) = -0.18
The volume doesn't change, so we can replace the concentration by the number of moles. The acid will react with the A⁻ and will form HA so:
A⁻f = 0.150 - nHCl
HAf = = 0.100 + nHCl
pKa - log((0.150 - nHCl)/(0.100 + nHCl)) - pKa - log(0.150/0.100) = -0.18
log((0.150 - nHCl)/(0.100 + nHCl)) - log(0.150/0.100) = - 0.18
log((0.150 - nHCl)/(0.100 + nHCl))/(0.150/0.100)= -0.18
((0.150 - nHCl)/(0.100 + nHCl))/(0.150/0.100)= [tex]10^{-0.18}[/tex]
((0.150 - nHCl)/(0.100 + nHCl))/(0.150/0.100) = 0.6607
(0.150 - nHCl)/(0.100 + nHCl) = 0.6607*(0.150/0.100)
(0.150 - nHCl)/(0.100 + nHCl) = 0.9910
(0.150 - nHCl) = 0.9910*(0.100 + nHCl)
0.150 - nHCl = 0.0991 + 0.991nHCl
1.991nHCl = 0.0509
nHCl = 0.025 mol
To change the pH of the buffer by 0.18 units, the number of moles of HCl to be added is 0.025 moles.
What is a buffer?
A buffer is a solution which resists changes to its pH when small amounts of strong acids or base are added to it.
The equation of the dissociation of NaH₂PO₄ is given below:
NaH₂PO₄ → Na⁺ + H₂PO₄⁻
From the equation, the number of moles of H₂PO₄⁻ equals to the number of moles of NaH₂PO₄
moles of H₂PO₄⁻ = 0.250 L * 1 mol/L
moles of H₂PO₄⁻ = 0.250 mol
Number of moles of NaOH that reacts = 0.150 L * 1 mol/L
moles of NaOH that reacts = 0.150 mol
Assuming all the base reacts with the acid H₂PO₄⁻ to form the conjugate base HPO₄²⁻
moles of H₂PO₄⁻ left unreacted = 0.250 - 0.150 = 0.100 moles
moles of HPO₄²⁻ formed = 0.150 moles
The pH of a buffer is calculated using the Henderson-Hasselbalch equation:
- pH = pKa + log[A⁻]/[HA]
Where;
- pKa = -logKa,
- [A⁻] is the conjugate base concentration (HPO₄²), and
- [HA] is the acid concentration (H₂PO₄⁻)
For a pH change of 0.18 units, the concentrations of A⁻ and HA will change too.
let [A⁻]ₓ and [HA]ₓ be the final concentrations.
The pH will decrease since an acid, HCl is added.
Thus;
- pH final - pH initial = -0.18
- (pKa + log[A⁻]ₓ/[HA]ₓ) - (pKa + log[A⁻]/[HA]) = -0.18
Since the total volume does not change, the concentration is equated to the number of moles.
The acid will react with the A⁻ and will form HA thus:
A⁻ₓ = 0.150 - nHCl
HAₓ = = 0.100 + nHCl
where;
- nHCl is number of moles of HCl
Substituting in the equation above
pKa - log{(0.150 - nHCl)/(0.100 + nHCl)} - pKa - log(0.150/0.100) = -0.18
log{(0.150 - nHCl)/(0.100 + nHCl)} - log(0.150/0.100) = -0.18
log{{(0.150 - nHCl)/(0.100 + nHCl) / (0.150/0.100)}} = -0.18
{(0.150 - nHCl)/(0.100 + nHCl)} / (0.150/0.100) = 0.6607
(0.150 - nHCl)/(0.100 + nHCl) = 0.6607 * 1.5
(0.150 - nHCl)/(0.100 + nHCl) = 0.991
(0.150 - nHCl) = 0.9910 * (0.100 + nHCl)
0.150 - nHCl = 0.0991 + 0.991nHCl
1.991nHCl = 0.0509
nHCl = 0.025 mol
Therefore, number of moles of HCl to be added is 0.025 moles.
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