Answer on Question #70767 – Chemistry – Other
Task:
What is the hydrogen ion concentration of 0.1 N boric acid.
A) 1.15 × 1 0 − 12 1.15 \times 10^{-12} 1.15 × 1 0 − 12 ;
B) 3.7 × 1 0 − 8 3.7 \times 10^{-8} 3.7 × 1 0 − 8 ;
C) 4.82 × 1 0 − 4 4.82 \times 10^{-4} 4.82 × 1 0 − 4 ;
D) 6.3 × 1 0 − 6 6.3 \times 10^{-6} 6.3 × 1 0 − 6 .
Solution:
The dissociation of boric acid in the first stage proceeds according to the reaction:
H 3 B O 3 − = H + + H 2 B O 3 − , H_3BO_3^- = H^+ + H_2BO_3^-, H 3 B O 3 − = H + + H 2 B O 3 − ,
For which the dissociation constant K d 1 = 5.8 × 1 0 − 10 K_{d1} = 5.8 \times 10^{-10} K d 1 = 5.8 × 1 0 − 10 ,
K d 1 = [ H 2 B O 3 − ] ⋅ [ H + ] [ H 3 B O 3 − ] = 5.8 × 1 0 − 10 . K_{d1} = \frac{[H_2BO_3^-] \cdot [H^+]}{[H_3BO_3^-]} = 5.8 \times 10^{-10}. K d 1 = [ H 3 B O 3 − ] [ H 2 B O 3 − ] ⋅ [ H + ] = 5.8 × 1 0 − 10 .
The second and third stages of dissociation of boric acid are neglected.
[ H + ] = K d 1 ⋅ C m = 6.3 × 1 0 − 6 . [H^+] = \sqrt{K_{d1} \cdot C_m} = 6.3 \times 10^{-6}. [ H + ] = K d 1 ⋅ C m = 6.3 × 1 0 − 6 .
Answer: D) 6.3 × 1 0 − 6 6.3 \times 10^{-6} 6.3 × 1 0 − 6 mol/dm³.
Answer provided by https://www.AssignmentExpert.com