Hydrofluoric acid is a weak acid with a K_a, = 6.8 times 10^4. A titration of 25

Question

Hydrofluoric acid is a weak acid with a K_a, = 6.8 times 10^4. A titration of 25.0 mL of 0.500 M hydrofluoric acid is carried out using 0.200 M NaOH. What is the initial pH of the hydrofluoric acid solution? Write the chemical reaction equation for this titration. 50.0 mL of 0.200 M NaOH is added to the initial HF solution. At what point in the titration does this represent (i.e., before, at, or after the equivalence point)? Support your answer with calculations and/or a logical explanation. What is the pH of this solution after the 50.0 mL of 0.200 M NaOH is added to the initial HF solution? Don't forget to account for dilution.

Explanation / Answer

HF------à H+ +F-

Ka= [H+] [F-]/[HF]

Let x= concentration of [H+] =[F-] at equilibrium

Ka= 6.8*10-4= x2/(0.5-x)

Since HF is weak 0.5-x can be approximated to 0.5

6.8*10-4= x2/0.5

Therefore x2= .00034, x=0.018439, pH= -log(0.014439)=1.73

When NaOH is added, the reaction is HF+NaOH----à NaF+H2O

1 mole of HF requires 1 mole of NaOH

Molesof HF taken= 0.5*25/1000=0.0125 moles

Molesof NaOH taken = 0.2*50/1000=0.01 moles

This is less than the moles of HF, this point is before equivalenece point

HF is excess and it is excess by= 0.0125-0.01=0.0025 moles

Volume after mixing = 50+25= 75ml =75/1000=0.075

Concentration of HF= 0.0025/0.075=0.0333M

Now x2/(0.033-x)= 6.8*10-4

when solved using excel , x =0.0044, pH= -log(0.0044)=2.35

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