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How much of 0.3 M ammonium hydroxide should be mixed with 30 mL of 0.2 M solution of ammonium chloride to give buffer solutions of pH 8.65 and 10 (pK_b = 4.75) ? |
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Answer» SOLUTION :For BASIC buffer solution, `PH=14-pK_b+"LOG""[Base]"/"[Salt]"` pH=14-4.75 +log `"[Base]"/"[Salt]"` or log `"[Base]"/"[Salt]"` = pH-9.25 (i)For suffer of pH=8.65 log `"[Base]"/"[Salt]"`=8.65-9.25 or `"[Base]"/"[Salt]"` =Antilog (-0.60)=0.25 or `([NH_4OH])/([NH_4Cl])=((0.3xxV)/1000)/((0.2xx30)/1000)`=0.25 `V=(0.25xx0.2xx30xx1000)/(0.3xx1000)` =5.02 ML Similarly for solution of pH =10 log `"[Base]"/"[Salt]"` =10-9.25 `"[Base]"/"[Salt]"` =antilog 0.75=5.62 `therefore ((0.3xxV)/1000)/((0.2xx30)/1000)=5.62` `therefore V=(5.62xx0.2xx30xx1000)/(0.3xx1000)` =12.94 mL |
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