Determine the pH of a buffer made from 250 ml of 0.50M HF and 150 ml of 0.75 M NaF.
Substitute into the HH equation. HF is the acid; NaF is the base.
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Calculate the ph of a buffer solution prepared from 0.25m HF and 0.5m CaF2
To determine the pH of a buffer, you need to consider the equilibrium reaction between the acid and its conjugate base. In this case, the acid is hydrofluoric acid (HF) and its conjugate base is fluoride ion (F-).
HF ⇌ H+ + F-
A buffer is a solution that resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base, or a weak base and its conjugate acid.
To calculate the pH of the buffer, we need to consider the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
First, let's determine the concentrations of HF and NaF:
- Volume of HF = 250 ml
- Concentration of HF = 0.50 M
- Moles of HF = Volume * Concentration = 250 ml * 0.50 M = 125 mmol
- Volume of NaF = 150 ml
- Concentration of NaF = 0.75 M
- Moles of NaF = Volume * Concentration = 150 ml * 0.75 M = 112.5 mmol
Since both HF and NaF dissociate completely, the moles of F- will be the same as the moles of NaF, and the moles of H+ will be the same as the moles of HF.
Now, let's calculate the molar concentrations of HF and F-:
- Volume of the final buffer = Volume of HF + Volume of NaF = 250 ml + 150 ml = 400 ml
- Molar concentration of HF = Moles of HF / Volume of the final buffer = 125 mmol / 400 ml = 0.3125 M
- Molar concentration of F- = Moles of F- / Volume of the final buffer = 112.5 mmol / 400 ml = 0.28125 M
Next, we need to find the pKa value for HF. The pKa value can be found in reference materials or online databases. For HF, the pKa value is approximately 3.2.
Now, we can substitute the values in the Henderson-Hasselbalch equation:
pH = 3.2 + log ([F-]/[HF])
= 3.2 + log (0.28125 M / 0.3125 M)
= 3.2 + log (0.9)
Calculating the log (0.9) using a calculator, we find:
log (0.9) ≈ -0.0458
Now, we substitute this value back into the equation:
pH ≈ 3.2 - 0.0458
pH ≈ 3.1542
Therefore, the pH of the buffer made from 250 ml of 0.50 M HF and 150 ml of 0.75 M NaF is approximately 3.1542.