Question

What would be the pH at the equivalence point of a 0.01M solution of strong acid titrated using 45mL of 0.02Mof a strong base? *

Answer 1

`pH = 9.39`

Step-by-step explanation:

We will first derive the volume of base to reach the equivalence point.

As we know that

`m_1V_1 = m_2 V_2`

Substituting the given values we get

`0.01 * V_1 = 0.02 * 45\\V_1 = 90`

Concentration of slat formed

`(0.02 * 45 )/(45 + 90) \\= 0.067`

H+ ion concentration is equal to

`\sqrt{(K_w * K_c )/(C) }`

Substituting the given values in above equation, we get -

`= \sqrt{(10^(-14) * 1.8* 10^(-5))/(0.067) } \\= 4.047 *10^(-10)`

`pH = -log [H^+]`

`pH = 9.39`

Answer 2

Step-by-step explanation:

45 mL = .045 L

.045 L of .02M base will contain .045 x .02 mole of base

= .0009 mole of base

acid = .01 mole

net acid = .01 - .0009

= .0091 mole

this mole has volume of .045L

concentration of acidic solution

= .0091 / .045

= .2022 M

= 2022 x 10⁻⁴ M

pH = - log 2022 x 10⁻⁴

= 4 - log 2022

= 4 - 3.3

= .7

pH = .7 .