See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

2 3 3 2 3 H CO NaOH NaHCO Na CO 10 15 0 0 0 0 5 5    pH = x = 10 + 5 log 5       = 10 After adding 2 4 H SO 3 2 3 2 3 NaHCO Na CO H H CO 5 5 10 0 5 0 0 5     pH = y = 6 + 5 log 5       = 6 x – y = 10 – 6 = 4