What is pH when 0.5 m HAC (acetic acid k_a = 1.8 times 10^-5) is allowed to go to equilibrium? HAC + H_2O = AC^- + H_3O^+

## Expert Answer

Answer

**HAC dissociates as:**

**HAC + H2O —–> H3O+ + Ac-**

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Question & Answer: What is pH when 0.5 m HAC (acetic acid k_a = 1.8 times 10^-5) is allowed to go to…..

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**0.5 0 0**

**0.5-x x. x**

**Ka = [H+][Ac-]/[HAC]**

**Ka = x*x/(c-x)**

**Assuming x can be ignored as compared to c**

**So, above expression becomes**

**Ka = x*x/(c)**

**so, x = sqrt (Ka*c)**

**x = sqrt ((1.8*10^-5)*0.5) = 3*10^-3**

**since c is much greater than x, our assumption is correct**

**so, x = 3*10^-3 M**

**so.[H+] = x = 3*10^-3 M**

**use:**

**pH = -log [H+]**

**= -log (3*10^-3)**

**= 2.52**

**Answer: 2.52**