Respuesta :
Explanation:
First, we will calculate the number of moles of HF as follows.
No. of moles of HF = volume × concentration of HF
= [tex]1.0 L \times 1.78 M[/tex]
= 1.78 mol
According to Henderson-Hasselbalch equation,
pH = [tex]pK_{a} + log (\frac{[NaF]}{[HF]})[/tex]
= [tex]-log K_{a} + log (\frac{\text{moles of NaF}}{\text{moles of HF}})[/tex]
Since, no change in the volume is taking place. Hence, putting the given values into the above formula as follows.
pH = [tex]-log K_{a} + log (\frac{\text{moles of NaF}}{\text{moles of HF}})[/tex]
[tex]3.35 = -log (7.2 \times 10^{-4}) + log (\frac{\text{moles of NaF}}{1.78})[/tex]
[tex]log (\frac{\text{moles of NaF}}{1.78})[/tex] = 0.208
[tex](\frac{\text{moles of NaF}}{1.78})[/tex] = 1.614
moles of NaF = 2.87 mol
Thus, we can conclude that moles of solid NaF present are 2.87 mol.
The moles of solid NaF present are 2.87 mol.
Calculation for moles of HF:
No. of moles of HF = volume × concentration of HF
No. of moles of HF= 1.0 L* 1.78 M
No. of moles of HF = 1.78 mol
According to Henderson-Hasselbalch equation,
pH = pKa + log ([NaF] / [HF])
pH = pKa + log ([Moles of NaF] / [Moles of HF])
There is no change in the volume is taking place.
On substituting the values:
pH = pKa + log ([Moles of NaF] / [Moles of HF])
3.35 = -log ([tex]7.2*10^{-4}[/tex]) + log ([Moles of NaF] / [1.78])
log ([Moles of NaF] / [1.78]) = 0.208
([Moles of NaF] / [1.78]) = 1.614
Moles of NaF = 2.87 mol
Thus, the moles of solid NaF present are 2.87 mol.
Find more information about pH here:
brainly.com/question/22390063
