JEE Main Chemistry question
The percentage error in [H⁺] made by neglecting the ionisation of water in 1.0×10⁻⁶ M NaOH is:
Answer: (A)
Solution
True [OH⁻]=(Cb+√(Cb²+4Kw))/2≈1.0099×10⁻⁶, vs. the naive 1.0×10⁻⁶ — about a 1% error.
The percentage error in [H⁺] made by neglecting the ionisation of water in 1.0×10⁻⁶ M NaOH is:
True [OH⁻]=(Cb+√(Cb²+4Kw))/2≈1.0099×10⁻⁶, vs. the naive 1.0×10⁻⁶ — about a 1% error.