Answer to Question #286329 in Physical Chemistry for Bally

Question

Topic:Acid and Bases

Q1

a) calculate the pH of the following solutions:

i) 0.002 mol dm^-3 KOH (2 marks)

ii) 0.10 mol dm^-3 C₆H₅COOH ( K of benzoin acid = 6.3 x 10^-5 mol dm^-3 ) ( 4 marks)

b) The pH of a 0.10 mol dm^-3 solution of a weak mono proton acid, HA , is 2.85. Determine the K of the acid. ( 2 marks)

c) A 500 cm³ solution containing 1.8g of a weak acid HA has a pH of 3.5. Calculate the molar mass of the acid, given that it has a value for K of 2.0 x 10 ^-6mol dm . ( 5 marks)

please answers all the questions with detailed explanation and steps 🙏🏾🙏🏾🙏🏾🙏🏾🙏🏾🙏🏾

Answer 1

1 )

a )( i )

0.002 M = 0.002 mol / dm³ KOH Solution

So, Concentration of OH^- = 0.002 = 2 × 10^-3 .

So, P^OH = -log 2× 10^-3 = 3 - log 2 = 2.7

So , P^H = 14 - 2.7 = 11.3

( a ) ( ii ) H⁺ = under root C . K_a

= under root 0.1 × 6.3 × 10^-5

= under root 6.3 × 10^-6 .

= 2.6 × 10^-3 .

P^H = - log 2.6 × 10^-3

P^H = 3 - log 2.6 = 2.6

( b ) p^H = 2.85 .

(H⁺)² = (10^-2.85)² = 0.1 × k

So, k = 10^--4.7 = 2 × 10^-5 .

( c ) p^H = 3.5

( H⁺ ) ² =( 10^-3.5 )² = c × k

10^-7 = ( 1.8 / M × 0.5 ) × 2 × 10^-6 .

M = molar mass of weak acid = 72 g / mol.