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Calculate the pH of the resultant mixtures : 10 mL 0.1 M H_2SO_4 + 10 mL 0.1 M KOH |
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Answer» SOLUTION :Millimol=mL x M In Initial MOL of `H_2SO_4` = 10 x 0.1 = 1.0 mol `H_2SO_4` In initial m mole of KOH = 10 x 0.1 = 1.0 m mol KOH Total volume of mix solution20 mL = 0.02 L Millimol of remaining `H_2SO_4` after neutralization `{:(H_2SO_4 + , 2KOH to , K_2SO_4 + , 2H_2O),("1 mol","2 mol" , "1 mol", "2 mol"):}` `therefore` (m mol of remaining `H_2SO_4` ) = (m mol `H_2SO_4` in initial ) -(`H_2SO_4` consumed in neutralization ) =(1 m mol- `1/2` m mol ) `H_2SO_4` =0.5 m mol `H_2SO_4` remaining =0.5/1000 mol `H_2SO_4` remaining Molarity of remaining , `H_2SO_4 =("mol"/"Total volume (L)")` `=(0.5xx1000)/(1000xx20)` `=0.5/20`=0.025 M `H_2SO_4` Remaining `[H^+]` in mixture = `2[H_2SO_4]` =2(0.025) = 0.05 PH = log `[H^+]` = log (0.05) =-(-1.3010)=1.3010 |
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