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

Answer:

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 .

Answer 2

Answer:

$pH = 9.39$

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

$\frac{0.02 * 45 }{45 + 90} \\= 0.067$

H+ ion concentration is equal to

$\sqrt{\frac{K_w * K_c }{C} }$

Substituting the given values in above equation, we get -

$= \sqrt{\frac{10^{-14} * 1.8* 10^{-5}}{0.067} } \\= 4.047 *10^{-10}$

$pH = -log [H^+]$

$pH = 9.39$