Determine the pH of a solution that is prepared by mixing 100.0 mL of 0.125 M NaOH with 145 mL of 0.125 M CH3COOH (pKa = 4.75). Report your answer to 2 decimal places.
Please help, I have no clue!!
1. Realize that NaOH and CH3COOH is the reaction of a strong base and a weak acid. Write and balance the equation. The products are CH3COONa and HOH.
2. Determine mols NaOH and mols CH3COOH. mols = L x M.
3. From the equation, determine mols CH3COONa formed. That is a salt. I haven't worked the problem but you PROBABLY will have an excess of CH3COOH which will make a buffer of CH3COOH and CH3COONa (a weak acid and its salt) and all of the NaOH will be used.
4. Then use the Henderson-Hasselbalch equation. pH = pKa + log [(base}/(acid)]
Show your work if you get stuck.
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To determine the pH of the solution, we can use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log(A-/HA)
where pH is the resulting pH, pKa is the negative logarithm of the acid dissociation constant (Ka) for the weak acid (in this case, acetic acid or CH3COOH), A- is the concentration of the conjugate base (acetate ion or CH3COO-) and HA is the concentration of the weak acid.
First, let's calculate the concentrations of CH3COOH (HA) and CH3COO- (A-):
Given that the solution was prepared by mixing 100.0 mL of 0.125 M NaOH with 145 mL of 0.125 M CH3COOH, we need to determine the number of moles of CH3COOH and CH3COO- in the solution.
Number of moles of CH3COOH = volume (L) × molarity (M)
= 0.145 L × 0.125 M
= 0.018125 moles
Number of moles of NaOH = volume (L) × molarity (M)
= 0.100 L × 0.125 M
= 0.0125 moles
NaOH reacts with CH3COOH in a 1:1 ratio, so the number of moles of CH3COOH that reacts with NaOH is 0.0125 moles.
The remaining moles of CH3COOH are given by:
Remaining moles of CH3COOH = initial moles - moles reacted with NaOH
= 0.018125 - 0.0125
= 0.005625 moles
Since the volumes of the two solutions are added, the total volume of the resulting solution is 100 mL + 145 mL = 245 mL = 0.245 L.
Concentration of CH3COOH (HA):
Concentration (M) = moles / volume (L)
= 0.005625 moles / 0.245 L
= 0.022959 M
Concentration of CH3COO- (A-):
Since NaOH and CH3COOH react in a 1:1 ratio, the number of moles of CH3COO- formed is also 0.0125 moles.
Concentration (M) = moles / volume (L)
= 0.0125 moles / 0.245 L
= 0.05082 M
Now plug these values into the Henderson-Hasselbalch equation:
pH = pKa + log(A-/HA)
= 4.75 + log(0.05082/0.022959)
Using a calculator, compute the value inside the logarithm:
log(0.05082/0.022959) = 0.857
Finally, add this value to the pKa and round the resulting pH value to 2 decimal places:
pH = 4.75 + 0.857
= 5.60
Therefore, the pH of the solution is 5.60.