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InterviewSolution
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A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures of acids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is `pH=pK_(a) + log. (["Salt"])/(["Acid"])` a similar equation is used for basic buffers. The pH of aqueous solution of single salts is calculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is `pH = 7-(1)/(2) pK_(b)-(1)/(2) log c` For weak acids and bases used by a chemist, data are given below: `K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5)` Also logarithmic values of some numbers are given below : log 1.8 = 0.2553, log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH value in each of the following cases. 100 mL of 0.05 M NaOH mixed with 100 ml of 0.10 M `CH_(3)CO OH` solutionA. 3.75B. 4.75C. 5.75D. 6.75 |
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Answer» Correct Answer - B 100 ml of 0.05 M NaOH = 5 mmol of NaOH 100 mL of 0. 1 M `CH_(3)CO OH= 10 ` mmol of `CH_(3)CO OH` 5 mmol of NaOH will react with 5 mmol of `CH_(3)CO OH` to form 5 mmol of `CH_(3)CO ON a` `CH_(3)CO OH ` left = 5 mmol, `CH_(3)CO ON a` formed = 5 mmol Total volume = 200 mL `:. [CH_(3)CO OH]=[CH_(3)CO ON a]` `=(5)/(200) M = 0.025 M` As it is a buffer of weak acid and its salt with strong base, `pH = pK_(a)+log. (["Salt"])/(["Base"])` `=-log (1.8xx10^(-5))+log 1` `=5- (0.255)=4.745` |
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