A solution is prepared by mixing 100 mL 0.50 M hydrazoic acid (HN 3 ), whose K a = 3.6 ×…
Chemistry · JEE Advanced · NTA Exams — Equilibrium
A solution is prepared by mixing 100 mL 0.50 M hydrazoic acid (HN3), whose Ka = 3.6 × 10–4, with 400 mL of 0.10M cyanic acid (HOCN), whose Ka = 8 × 10–4. Which of the following is (are) true regarding
- [H+] = 10–2 M
- \(\left[\mathrm{N}_{3}^{-}\right]\)= 3.6 × 10–3
- [OCN–] = 4.57 × 10–3
- [OH–] = 7.14 × 10–13
Answer
(A) [H + ] = 10 –2 M, (B) [ N _ 3 ^ - ] = 3.6 × 10 –3
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