Question

5 acetic acid solutions, 0.10M, 0.50M,1.00M, 3.00M, 6.00M.

(PL5) Calculate pH for a solution which has [H+] = 0.765M.

(PL6) Calculate [H+] for a solution with pH=3.65

A1) Calculate percent dissociation of 5 solutions. Show your work clearly for 0.10M solution.

(A2) Draw a graph of percent dissociation vs. concentration. Label everything.

(Q1) What is the theoretical pH of 1.00M acetic acid (Ka = 1.8x10^-5)

(Q2) What is the theoretical percent dissociation for Q1?

(Q3) What is the theoretical pH of 6.00M acetic acid (Ka = 1.8x10^-5)?

(Q4) What is the theoretical percent dissociation for Q3?

(Q5) Looking at your answers so far, what is the theoretical relationship between concentration and percent dissociation?

(Q6) When 0.050 M HF solution has pH of 2.40, what is the percent dissociation?

(Q7) Compare your experimental pH / percent dissociation with theoretical pH / percent dissociation calculated here. Briefly discuss.

Answer 1

Calculate pH for a solution which has [H+] = 0.765M.

Solution :-

Formula to calculate the pH

pH= -log [H^+]

so lets put the given concentration of the H+ in the formula and calculate the pH

pH= -log [0.765]

pH= 0.116

Calculate [H+] for a solution with pH=3.65

Solution :-

Using the given pH value we can find the [H^+] concentration as follows

pH= -log [H^+]

Rearranging the formula we get

[H^+] = antilog (- pH)

So lets put the value of the pH in the formula and calculate the [H^+]

[H^+] = antilog(-pH)

           =10^(-3.65)

           = 2.24*10^-4 M

Therefore the concentration of the [H^+] = 2.24*10^-4 M