2NaHCO3 + H2SO4 ==> Na2SO4 + 2CO2 + 2H2O
millimoles NaHCO3 = mL x M = 10.00 x 0.100 = 1.000
millimoles H2SO4 = 25.0 x 0.078 = 1.95
mmoles H2SO4 needed = (1 molH2SO4/2 mol NaCHO3) = 0.500 mmol H2SO4 needed. You have that much; therefore, NaHCO3 is the limiting reagent (LR) and H2SO4 is the excess reagent (ER).
So ALL of the NaHCO3 will be used and you will have how much leftr of the original H2SO4? That's 1.95 mmoles initially - 0.500 mmoles used = 1.45 mmoles H2SO4 left in a volume of 10 + 25 = 35 mL.
(H2SO4) = mmoles/mL = 1.45/35 mL = 0.041 M
The first ionization of H2SO4 is 100%. The second one has a k2 = 0.012. Total will be the sum of 0.041 from the first H^+ plus what we get from k2 for H2SO4. Here is the first one.
..................H2SO4 ==> H^+ + HSO4^-
I..................0.041...........0............0
C..................-0.041........0.041........0.041
E...................0................0.041..........0.041
.................HSO4^- ===> H^+ + [SO4]^2-
I..................0.041............0.041.......0
C...................-x.................+x...........+x
E................0.041-x.......0.041+x.........+x
k2 = (H^+)(SO4^2-)/(HSO4^-)
0.012 = (0.041+x)(x)/(0.041-x). Solve for x. Evaluate 0.041+x and plug in for H^+ in pH = -log(H^+). Post your work if you get stuck.
If 10.00ml of a 0.100M NaHCO3 solution is added to 25ml of 0.078M H2SO4 what will be pH of solution
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1 answer