
J.R. S. answered 11d
Ph.D. University Professor with 10+ years Tutoring Experience
a) KOH + HNO3 ==> KNO3 + H2O
moles KOH present = 60.0 ml x 1 L/1000 ml x 0.125 mol/L = 0.0075 moles
moles HNO3 needed = 0.0075 b/c they react in a 1:1 mole ratio as seen in the balanced equation
Volume of 0.150 M HNO3 needed: (x L)(0.150 mol/L) = 0.0075 mol and x = 0.05 L = 50 mls
NOTE: because these react in a 1:1 mole ratio, you could have used V1M1 = V2M2 as follows:
(x ml)(0.150 M) = (60 ml)(0.125 M) and x = 50 mls. But be careful using this method as it really only works if the acid and base react in a 1:1 mole ratio.
b) 3KOH + H3PO4 ==> K3PO4 + 3H2O (note that the acid and base DO NOT react in a 1:1 mole ratio)
moles KOH present = 60.0 ml x 1 L/1000 ml x 0.125 mol/L = 0.0075 moles
moles H3PO4 needed = 0.0075 mol KOH x 1 mol H3PO4 / 3 mol KOH = 0.0025 moles
Volume of 0.200 M H3PO4 needed: (x L)(0.200 mol/L) = 0.0025 moles and x = 0.0125 L = 12.5 mls