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The pH of a solution obtaine by mixing `50 mL` of `0.4 N HCl` and `50 mL` of `0.2 N NaOH` isA. log 2B. `1.0`C. `-log (0.05)`D. `2.0` |
Answer» Correct Answer - B Acid left unconsumed `=(50 xx 0.2)/( 0.4)` `=25 mL ` fo `0.4 N` New normality of resulting solution `=25 xx 0.4 -= 100xx N =0.1 N` `pH =-log [10^(-1)]=1` |
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