HNO3 + KOH ==> KNO3 + H2O
a)HNO3 is 100% ionized; therefore, pH = -log(H^+) = -log(HNO3) = ?
b)
mols HNO3 = M x L = ?
mols KOH added = ?
mols HNO3-mols KOH = mols HNO3 remaining.
M HNO3 remaining = mols HNO3/total volume. Don't forget total volume will be amount HNO3 you started with + KOH volume added.
c.
mols HNO3 = M x L
mols KOH needed = mols HNO3 (see the coefficients in the balanced equation.)
mols KOH = M x L. You know mols and M, solve for L. Convert to mL if needed.
A titration is performed by adding 0.124 M KOH to 40 mL of 0.159 M HNO3.
a) Calculate the pH before addition of any KOH.
b) Calculate the pH after the addition of 10.26, 25.65 and 50.29 mL of the base.(Show your work in detail for one of the volumes.)
c) Calculate the volume of base needed to reach the equivalence point.
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