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If the equilibrium constant for the reaction of weak acid HA with strong base is `10^(9)`, then pH of `0.1M` Na A is:A. 5B. 9C. 7D. 8 |
Answer» Correct Answer - B `HA +NaOH hArrNaA +H_(2)O` `K=10^(9)` `:.` For `NaA +H_(2)O hArr NaOH +HA,K=10^(-9)` Initial 0.1 M At eqm `(0.1 -alpha)M " "alpha M " "alphaM` `([NaOH][HA])/([NaA])=10^(-9)` `(alpha^(2))/(0.1-alpha)=10^(-9)` As `alpha` is very small , `0.1 -alpha ~~0.1` `alpha^(2)=10^(-9)xx0.1` `alpha=10^(-5)` `[OH^(-)]=[NaOH]=10^(-5)M` ltbr. `pH =14 0pOH =14-5=9` |
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