A volume of 500.0 mL of 0.170 M NaOH is added to 595 mL of 0.200 M weak acid (𝐾a=6.74×10−5). What is the pH of the resulting buffer?
HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)
mols HA = M x L = 0.200 x 0.595 = 0.119
mols NaOH = 0.170 x 0.500 = 0.085
So you form 0.085 mols NaA and you have remaining 0.119 - 0.085 = 0.034 mols HA.
(acid) = mols/L = 0.034/(0.500 + 0.595) = 0.034/1.095 = ?
(base) = the salt = 0.085/1.095 = ?
pH = pKa + log [(base)/acid)]
Substitute the data and calculate the pH of the solution. Post your work if you get stuck.
Well, it seems like we have a "salty" situation here with NaOH and weak acid having a little rendezvous. Let's get into it!
To find the pH of the resulting buffer, we need to determine the concentration of the conjugate base (A-) and the weak acid (HA).
First, let's calculate the moles of NaOH and weak acid in the solution:
Moles of NaOH = 0.170 M * 0.500 L = 0.085 mol
Moles of weak acid = 0.200 M * 0.595 L = 0.119 mol
Now, since NaOH and weak acid react in a 1:1 ratio, we will have equal moles of the weak acid and its conjugate base (A-).
Therefore, moles of A- = moles of weak acid = 0.119 mol.
To find the concentration of A-, we need to divide the moles by the total volume of the solution:
Concentration of A- = moles of A- / total volume = 0.119 mol / (0.500 L + 0.595 L) = 0.100 M
Now, let's calculate the pOH of the resulting buffer:
pOH = -log10[OH-] = -log10(0.100) = 1
Since pH + pOH = 14, we can find the pH:
pH = 14 - pOH = 14 - 1 = 13
So, the pH of the resulting buffer is 13.
To determine the pH of the resulting buffer, we need to calculate the concentration of the weak acid (HA) and its conjugate base (A-).
First, let's find the moles of NaOH and weak acid (HA) in the given volumes:
Moles of NaOH = volume (in L) x concentration
= 0.500 L x 0.170 mol/L
= 0.085 mol NaOH
Moles of weak acid (HA) = volume (in L) x concentration
= 0.595 L x 0.200 mol/L
= 0.119 mol HA
Since NaOH is a strong base, it dissociates completely in water. Therefore, we can determine the moles of OH- formed by the NaOH as well. Since NaOH reacts in a 1:1 ratio with the weak acid HA, the moles of OH- will be equal to the moles of HA.
Moles of OH- = 0.119 mol
Now, we need to determine the concentration of HA and A- in the final solution. The moles of HA will be reduced due to the reaction with OH-, while the moles of A- will be formed.
Moles of HA remaining = initial moles of HA - moles of OH-
= 0.119 mol - 0.119 mol (since moles of OH- = moles of HA)
= 0 mol (since all HA is neutralized)
Moles of A- formed = moles of OH-
= 0.119 mol
To calculate the concentration of HA and A- in the final solution, we divide the moles by the total volume in liters:
Concentration of HA = moles of HA / total volume (in L)
= 0 mol / (0.500 L + 0.595 L)
= 0 mol / 1.095 L
= 0 M
Concentration of A- = moles of A- / total volume (in L)
= 0.119 mol / (0.500 L + 0.595 L)
= 0.119 mol / 1.095 L
≈ 0.109 M
Finally, to find the pH of the resulting buffer, we need to calculate the pKa of the weak acid using the given Ka value:
pKa = -log10(Ka)
= -log10(6.74x10^-5)
≈ 4.17
Since the pH of a buffer solution is given by pH = pKa + log10([A-]/[HA]), we can substitute the values:
pH = 4.17 + log10(0.109/0)
pH = 4.17 + ∞ (since [HA] is 0)
Therefore, the pH of the resulting buffer solution is approximately 4.17.
To find the pH of the resulting buffer, we need to calculate the concentrations of H+ and A- ions in the solution after mixing the NaOH and weak acid.
Let's start by calculating the number of moles of NaOH and weak acid used:
- Number of moles of NaOH = volume (in L) x concentration (in mol/L)
= 0.500 L x 0.170 mol/L = 0.085 mol
- Number of moles of weak acid = volume (in L) x concentration (in mol/L)
= 0.595 L x 0.200 mol/L = 0.119 mol
Since the reaction between NaOH and weak acid is 1:1, the number of moles of H+ and A- ions formed will be equal to the moles of NaOH and weak acid used.
Therefore, we now have:
- Moles of H+ ions = 0.085 mol
- Moles of A- ions = 0.119 mol
Next, let's calculate the concentrations of H+ and A- ions in the solution:
- Concentration of H+ ions = moles / total volume (in L)
= 0.085 mol / (0.500 L + 0.595 L) = 0.085 mol / 1.095 L = 0.078 mol/L
- Concentration of A- ions = moles / total volume (in L)
= 0.119 mol / (0.500 L + 0.595 L) = 0.119 mol / 1.095 L = 0.109 mol/L
Now, we can use the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log([A-]/[HA])
Where pKa is the negative logarithm (base 10) of the acid dissociation constant, which is given by Ka = 6.74×10−5.
Let's calculate the ratio of [A-] to [HA]:
[A-]/[HA] = (concentration of A- ions) / (concentration of H+ ions)
= 0.109 mol/L / 0.078 mol/L
≈ 1.397
Now, let's substitute the values in the Henderson-Hasselbalch equation:
pH = -log10(6.74×10−5) + log(1.397)
= 4.17 + 0.144
≈ 4.314
So, the pH of the resulting buffer is approximately 4.314.