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In which volume ratio `NH_(4)Cl` and `NH_(4)OH` solutions (each 1 M ) should be mixed to get a buffer solution of pH 9.80 ? `(pK_(b) "of" NH_(4)OH=4.74)`A. `1:2.5`B. `2.5:1`C. `1:3.5`D. `3.5:1` |
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Answer» Correct Answer - C Suppose `V_(1) ` mL of 1 M `NH_(4)Cl` solution is mixed with `V_(2)` mL of 1 M `NH_(4)OH` solution . `V_(1)` mL of 1 M `NH_(4) Cl = V_(1)` millimole `V_(2)` mL of 1 M `NH_(4)OH=V_(2)` millimole `V_(2)` mL of 1 M `NH_(4) OH = V_(2)` millimole pH = 9.80. Hence, pOH = 14 - 9.80 = 4.20 Now, pOH = `pK_(b) + log .(["Salt"])/(["Base"])` `:. 4.20 = 4.74 + log .(V_(1))/(V_(2))` or `log. (V_(1))/(V_(2))= - 0.54 = bar(1) . 46 :. (V_(1))/(V_(2))=0.29 = (1)/(3.5)` |
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