JEE Main Chemistry question

Chemistry Equilibrium JEE Main

The [OH⁻] in 100 mL of 0.015 M HCl(aq.) is:

(A)5×10⁻¹² M
(B)3×10⁻¹⁰ M
(C)6.7×10⁻¹³ MCorrect
(D)2.0×10⁻⁹ M

Answer: (C)

Solution

[OH⁻]=Kw/[H⁺]=10⁻¹⁴/0.015=6.7×10⁻¹³ M.

Question ID VK-136379