Question

**Calculate the pH at 25**°**C of a
0.075** *M* **solution of a weak acid that
has** *K*_{a} **= 1.3**
×**10**^{−5}**.**

Answer #1

**Let the weak acid be written as HA**

**HA dissociates as:**

**HA -----> H+ + A-**

**7.5*10^-2 0 0**

**7.5*10^-2-x x x**

**Ka = [H+][A-]/[HA]**

**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 ((1.3*10^-5)*7.5*10^-2) =
9.874*10^-4**

**since x is comparable c, our assumption is not
correct**

**we need to solve this using Quadratic
equation**

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

**1.3*10^-5 = x^2/(7.5*10^-2-x)**

**9.75*10^-7 - 1.3*10^-5 *x = x^2**

**x^2 + 1.3*10^-5 *x-9.75*10^-7 = 0**

**This is quadratic equation (ax^2+bx+c=0)**

**a = 1**

**b = 1.3*10^-5**

**c = -9.75*10^-7**

**Roots can be found by**

**x = {-b + sqrt(b^2-4*a*c)}/2a**

**x = {-b - sqrt(b^2-4*a*c)}/2a**

**b^2-4*a*c = 3.9*10^-6**

**roots are :**

**x = 9.809*10^-4 and x = -9.939*10^-4**

**since x can't be negative, the possible value of x
is**

**x = 9.809*10^-4**

**so.[H+] = x = 9.809*10^-4 M**

**use:**

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

**= -log (9.809*10^-4)**

**= 3.01**

**Answer: 3.01**

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