1 × 10–5 M AgNO3 is added to 1 L of saturated — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

10–5M AgNO3 in saturate AgBr Ksp = (s + 10–5) (s) = s(10–5) = 4.9 ×10–13 s = 4.9 × 40–8 solution has [Ag+] = 10–5   – NO = 10–5 [Br–] = 5 ×10–8  = C K  KTotal = C1  + 1000 C2  + 1000 C3   10 10 5 – 3 –    + 10 10 5 – 3 –    + 10 10 8 – 3 –    6 × 10–7 + 7 × 10–7 + 40 ×10–10 = 13.04 × 10–7 Scm–1  13.04 ×10–5Sm–1 13040 × 10–8 Sm–1 = 13040