# chemistry

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calculate the PH during the titration of 50.00 ml of 0.300 M HNO3 with 0.600 M KOH after 0, 15.50, 25.00, and 40.00 ml of KOH have been added. Graph the titration curve.

Ph=

• chemistry - ,

calculate the Ph in the titration of 50.00 ml of 0.100 M ammonia with 0.100 M HCl after 0,24.00,50.00, and 74.00 ml acid have been added

Ka((NH3)= 1.8x10^(-5)

PH= ............ when 0 ml added.

PH= ............ when 24.00 ml added

PH= ............ when 50.00 ml added

PH= ............ when 74.00 ml added

• chemistry - ,

millimoles HNO3 = mL x M = 50.00 x 0.3M = 15 millimoles.
mmoles KOH = mL x 0.6M =
mL = 0, 0mmoles KOH
mL = 15.5, 9.3 mmoles
mL = 25.0, 15 mmoles
mL = 40.0, 24 mmoles
I'll do the 15.5 or 9.3 mmoles KOH added.
..........HNO3 + KOH ==> KNO3 + H2O
initial...15......0........0......0
change...-9.3...-9.3.......9.3.....9.3
equil....5.7......0.........9.3....9.3

You can see that you have a solution of HNO3 with KNO3 (a salt that does not change the pH of the solution upon hydrolysis). So (H^+) = (HNO3) so convert H^+ to pH.
At the equivalence point (mmoles acid = mmoles base), the pH is determined by the hydrolysis of the salt. An ICE chart will show you how to go about that.
After the equivalence point you will have an excess of the base; convert to OH^-, pOH, and pH.

For the weak base and strong acid, do the same kind of thing (ICE chart etc) but in cases where the weak base is present you must go through an ICE chart for a weak base. At all points between the beginning and the equivalence point, you have a buffered solution and you should use the Henderson[-Hasselbalch equation.

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