100 mL of buffer solution that is 0.15M HA (Ka= 6.8x10-5) and 0.20M NaA is mixed with 21mL of 0.35M NaOH. What is the pH of the resulting solution?
NaOH will react with HA.
Calculate moles, then:
0.00735 mole of NaOH+ 0.015 NaA = x : new amount of NaA
0.020 HA-0.00735 moles of NaOH=y: new amount of HA
-log(KA)=pka
pH=pka+log[x/y]
To find the pH of the resulting solution, we need to calculate the concentration of the conjugate acid and conjugate base after the addition of NaOH.
Let's start by calculating the number of moles of HA and NaA in the 100 mL buffer solution:
Number of moles of HA = (0.15 M) x (100 mL) / 1000 = 0.015 moles
Number of moles of NaA = (0.20 M) x (100 mL) / 1000 = 0.020 moles
Next, we need to calculate the number of moles of NaOH added:
Number of moles of NaOH = (0.35 M) x (21 mL) / 1000 = 0.00735 moles
Now, let's determine the limiting reactant. Since NaOH reacts with HA to form water and NaA, we compare the moles of NaOH to moles of HA:
Moles of NaOH : Moles of HA = 0.00735 moles : 0.015 moles
≈ 0.49 : 1
Since the ratio is less than 1, NaOH is the limiting reactant. This means that all of the NaOH will react with HA, and there will be some HA left over.
The moles of NaOH that react with HA are equal to the initial moles of NaOH added (0.00735 moles). This means we have 0.015 - 0.00735 = 0.00765 moles of HA remaining.
Now, let's calculate the moles of NaA formed:
Moles of NaA formed = moles of NaOH added = 0.00735 moles
Given that we have a total volume of 121 mL (100 mL buffer + 21 mL NaOH), we can find the final concentrations of HA and NaA:
Concentration of HA = moles of HA remaining / total volume
= 0.00765 moles / 0.121 L
≈ 0.063 M
Concentration of NaA = moles of NaA formed / total volume
= 0.00735 moles / 0.121 L
≈ 0.061 M
Now, we can use the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log10(concentration of A- / concentration of HA)
= -log10(Ka) + log10(0.061 M / 0.063 M)
Using the given Ka value (6.8x10^-5), we can substitute the values and calculate the pH:
pH = -log10(6.8x10^-5) + log10(0.061 M / 0.063 M)
pH = 4.64 + log10(0.968)
pH ≈ 4.64 + (-0.014)
pH ≈ 4.626
Therefore, the pH of the resulting solution is approximately 4.626.