How can you prepare a buffer solutions of pH 9. You are provided with 0.1 M NH_4OH solution and ammonium chloride crystals. (ii) What volume of 0.6 M sodium formate solution is required to prepare a buffer solution of pH 4.0 by mixing it with 100 ml of 0.8 M formic acid .

How can you prepare a buffer solutions of pH 9. You are provided with

1.	How can you prepare a buffer solutions of pH 9. You are provided with 0.1 M NH_4OH solution and ammonium chloride crystals. (ii) What volume of 0.6 M sodium formate solution is required to prepare a buffer solution of pH 4.0 by mixing it with 100 ml of 0.8 M formic acid .
Answer» Solution :`pOH=pK_b+log([salt t])/([base])` We KNOW that PH+pOH=14 `therefore9+pOH=14` `impliespOH=14-9=5` `5=4.7+log""([NH_4Cl])/([NH_4OH])` `0.3=log".([NH_4Cl])/0.1` `([NH_4Cl])/0.1`= antilog of(0.3) `[NH_4Cl]=0.1M times 1.995` `=0.1995M` `=0.2M` Amount of `NH_4Cl` required to prepare 1 litre 0.2 M solution = Strength of `NH_4Cl times` molar mass of `NH_4Cl` =`0.2 times 53.5` =10.70g 10.70g ammonium chloride is dissolved in water and the solution is MADE up to one litre to get 0.2 M solution. On mixing equal volume of the given `NH_4OH` solution and the PREPARED `NH_4Cl` solution will give a buffer solution with required pH value (pH=9) (II) `pH=pK_a+log".([sal t])/([acid])` `4=3.75+log""([sodium formate])/([formic acid])` [Sodium formate]= number of moles of HCOONa `=0.6 times V times 10^-3` [formic acid]=number of moles of HCOOH `=0.8 times100 times10^-3` `=80 times10^-3` `4=3.75+log""(0.6V)/80` antilog of 0.25=`(0.6V)/80` `0.6V=1.778 times80` `=1.78 times80` `=142.4` `V=(142.4mL)/0.6=237.33mL`