Part II . pH of a series of hydrochloric acid solutions. Obtain 10.0 mL of 0.10

Question

Part II. pH of a series of hydrochloric acid solutions.

Obtain 10.0 mL of 0.10 M hydrochloric acid Predict the value of the pH. Measure the pH with the pH meter. Record the value.

Take 1.00 mL of the 0.10 M HCl(aq) in the previous step, and put it in another clean beaker. Add 9.00 mL of deionized water and stir. What is the new concentration of HCl(aq)? Predict the pH. Measure the pH. Record the value.

Add 90.0 mL of deionized water to the diluted HCl in step 2. What is the new concentration of HCl(aq)? Predict the pH. Measure the pH. Record the value.

Part III. pH of acetic acid solutions. Remember, acetic acid is a weak acid, with a Ka = 1.8 x 105.

Obtain 10.0 mL of 0.10 M acetic acid and place in a clean dry 50.0 mL beaker. Predict the value of the pH. Measure the pH with the pH meter. Record the value.

Take 1.00 mL of the 0.10 M CH3COOH(aq) in the previous step, and put it in another clean beaker. Add 9.00 mL of deionized water and stir. What is the new concentration of CH3COOH(aq)? Predict the pH. Measure the pH. Record the value.

Add 90.0 mL of deionized water to the diluted CH3COOH(aq) in step 2. What is the new concentration of CH3COOH(aq)? Predict the pH. Measure the pH. Record the value.

Explanation / Answer

Solved the complete Part 2 with multiple steps, post multiple question to get the remaining answers

Part 2

a) HCl is a strong acid, hence it will dissociate completely forming H+ and Cl-

pH = -log[H+] = -log[0.10] = 1

b) Adding 1 mL HCl to 9 mL of water

Final molarity of HCl = Initial Molarity of HCl * (Volume of HCl/Volume of HCl + Volume of water)

=> 0.1 * 1/(1+9) = 0.01M

pH = -log[H+] = -log(0.01) = 2

c) Initially we have 10mL of 0.01M HCl and this solution we are adding 90mL of H2O

Final molarity of HCl = Initial Molarity of HCl * (Volume of HCl/Volume of HCl + Volume of water)

=> 0.01 * 10/(10+90) = 0.001M

pH = -log[H+] = -log(0.001) = 3

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