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buffer solution pH question?
0.10 mol of solid sodium hydrogen carbonate and 0.20 mol of solid sodium carbonate are
dissolved in the same beaker of water, transferred to a volumetric flask and made to
250.0 mL. The Ka for HCO3– is 4.7 x 10^–11.
What is the pH of the resulting buffer?
Another additional detail -> what will be the pH of solution after i add 20mL of 0.05M hydrochloric solution to the original solution ? (Original solution is the buffer solution made above
1 Answer
- ?Lv 41 decade agoFavorite Answer
In the solution
HCO3- (aq) <--Ka--> CO3 2-(aq) + H+(aq)
Ka=([CO3 2-(aq) ][H+(aq)])/[HCO3-(aq)]=4.7 x 10^–11
logKa=log{([CO3 2-(aq) ][H+(aq)])/[HCO3-(aq)]}
equation can be rearrange in to
-log[H+]=-loKa+log[CO32-]/[HCO3 -]
pH=pKa+log[CO32-]/[HCO3 -]
[CO3 -]=0.2 mol/250 mL=0.0008 M:[HCO32-]=0.10 mol/250mL=.0004M
pH=-log4.7 x 10^–11+log[0.0008 M]/[.0004M]=10.33+0.30=10.63