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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 ofacids/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 aqueoussolution of single salts iscalculated 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 valuein each of the following cases. 100 mL of 0.05 M NH_(4)OH mixed with 100 mL of 0.10 M HCl solution |
Answer» <html><body><p>1.6<br/>2.6<br/>3.6<br/>4.6</p>Solution :100 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> of 0.05 M `NH_(4)OH = 5`mmol of `NH_(4)OH` <br/> 100 mL of 0.<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> M <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> = 10 mmol of HCl <br/> 5 mmol of `NH_(4)OH ` will react with5 mmol of HCl to <a href="https://interviewquestions.tuteehub.com/tag/form-996208" style="font-weight:bold;" target="_blank" title="Click to know more about FORM">FORM</a> 5 mmol of `NH_(4)Cl` <br/> HCl left = 5 mmol . Volume of solution = 200 mL<br/> `:. [HCl]=(5)/(200) = 0.025M`, <br/> `[H^(+)] =0.025 M`, <br/> `pH = - <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> (0.025) = 1.602`</body></html> | |