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To a 100 mL of 0.1 M weak acid HA solution 22.5 mL of 0.2 M solution of NaOH are added.Now, what volume of 0.1 M NaOH solution be added into above solution, so that pH of resulting solution be 4.7: [Given :(K_b(A^(-))=5xx10^(-10)] |
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Answer» Solution :`HA+NaOHtoNaA+H_2O` `t=0 " " 10 " " 4.5 " " 0 " " 0` t=50% `" " 5 " " 0 " " 5` 5 mL NAOH is required |
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