Asked by Zac
50.0mL of 0.01M Ba(OH)2 is added to 150.0mL of 0.01M HNO3. What is the new pH of the solution?
Answers
Answered by
DrBob222
50 mL x 0.01M Ba(OH)2 = 0.5 millimoles.
150 mL x 0.01 M HNO3 = 1.5 mmols.
....Ba(OH)2 + 2HNO3 ==> Ba(NO3)2 + 2H2O
I...0..........1.50........0.........0
added 0.5...............................
C....-0.5......-1.00......0.5........
E.......0.......0.5.........0.5........
(HNO3) = 0.5 millimols/200 mL = 0.0025 M
I work in millimoles when there are so many zeros. For example, for the Ba(OH)2 part.
mols = M x L = 0.01 x 0.050 = 0.0005 mol.
But millimols = 50 x 0.01 = 0.500.
You can change to mols in the above if you wish by dividing millimoles by 1000.
150 mL x 0.01 M HNO3 = 1.5 mmols.
....Ba(OH)2 + 2HNO3 ==> Ba(NO3)2 + 2H2O
I...0..........1.50........0.........0
added 0.5...............................
C....-0.5......-1.00......0.5........
E.......0.......0.5.........0.5........
(HNO3) = 0.5 millimols/200 mL = 0.0025 M
I work in millimoles when there are so many zeros. For example, for the Ba(OH)2 part.
mols = M x L = 0.01 x 0.050 = 0.0005 mol.
But millimols = 50 x 0.01 = 0.500.
You can change to mols in the above if you wish by dividing millimoles by 1000.
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