Is 50.0mL of 1.0 MHF and 50.0 mL of 1.0M NaOH a buffer solution with a pH of 3?
No. Neither a buffered solution nor a pH of 3.
u dr bob, explain further
dunt just say NO, ok
To determine whether the given solution is a buffer with a pH of 3, we need to calculate the concentration of the conjugate acid and the conjugate base and then use the Henderson-Hasselbalch equation.
1. First, let's calculate the concentration of the conjugate acid (HF) and the conjugate base (F-) in the solution.
Given:
Volume of 1.0MHF = 50.0 mL = 0.050 L
Volume of 1.0M NaOH = 50.0 mL = 0.050 L
Concentration of HF (conjugate acid):
Concentration = moles/volume
Moles = concentration × volume
Moles of HF = 1.0 M × 0.050 L = 0.050 moles
Concentration of F- (conjugate base):
Since NaOH is a strong base and fully ionizes in water, the concentration of OH- is equal to the concentration of NaOH.
Moles of NaOH = 1.0 M × 0.050 L = 0.050 moles
The concentration of F- = moles of NaOH = 0.050 moles
2. Now let's use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log10([A-]/[HA])
Since HF is a weak acid (Ka for HF = 6.8 × 10^-4), we can use the pKa value of approximately 3.18 (rounded from -log10(6.8 × 10^-4)).
pH = 3.18 + log10([F-]/[HF])
Substituting the values we calculated earlier:
pH = 3.18 + log10(0.050/0.050)
Simplifying:
pH = 3.18 + log10(1)
Taking the logarithm of 1 (base 10) is 0:
pH = 3.18 + 0
pH = 3.18
Therefore, the pH of the given solution is 3.18. Since it is not exactly pH 3, the solution is not a buffer with a pH of 3.