62 5. To what volume must each of the following be diluted to give the concentra
ID: 704907 • Letter: 6
Question
62 5. To what volume must each of the following be diluted to give the concentration called for? (300ml) a. 10.0ml of 6.00N HNOs to give 0.200N solution b. 9.00ml of concentrated H2SO, (18.0M) to give a 0.300M solution (540ml) 6. 24.60ml of 0.185N HNO3 is titrated with 27.35ml of KOH solution. What is the normality of the KOH? (0.166N) 7. The molar mass of phosphoric acid is 98.0 grams. What is its equivalent weight in each of the following reactions? a. H3PO4 + Al(OH)3 AIPo, + 3H20 b. HsPo, + NaOH-> NaH2PO4 + H2O c. H3PO4 + 2NaOH-+ Na2HPO4 + 2H2O (32.7g) (98.0g) (49.0g) (32.7g) (98.0g) e. H3POs+Ca(OH)2-Ca(H2POa)2+2H2OExplanation / Answer
5.
a. Initial Final
N1 = 6N N2 = 0.2N
V1 = 10ml V2 =
N1V1 = N2V2
V2 = N1V1/N2
= 6*10/0.2 = 300ml >>>>answer
b. Initial Final
M1 = 18M M2 = 0.3M
V1 = 9ml V2 =
M1V1 = M2V2
V2 = M1V1/M2
= 18*9/0.3 = 540ml >>>>answer
6. HNO3(aq) + KOH(aq) ---------> KNO3(aq) + H2O(l)
HNO3 KOH
N1 = 0.185M N2
V1 = 24.6ml V2 = 27.35ml
N1V1 = N2V2
N2 = N1V1/v2
= 0.185*24.6/27.35 = 0.166N >>>answer
7.
a. H3Po4 + Al(OH)3 -----> AlPO4 + 3H2O
Equivalent weight = molar mass of acid/number of hydrogen atoms are replaced
= 98/3 = 32.7g
b. H3Po4 + NaOH ----------> NaH2Po4 + H2O
Equivalent weight = molar mass of acid/number of hydrogen atoms are replaced
= 98/1 = 98g
c. H3Po4 + 2NaOH ---------> Na2HPO4 + 2H2O
Equivalent weight = molar mass of acid/number of hydrogen atoms are replaced
= 98/2 = 49g
d.
H3Po4 + 3NaOH ---------> Na3PO4 + 2H2O
Equivalent weight = molar mass of acid/number of hydrogen atoms are replaced
= 98/3 = 32.7g
e. H3PO4 + Ca(OH)2 -----------> Ca(H2Po4)2 +2H2O
Equivalent weight = molar mass of acid/number of hydrogen atoms are replaced
= 98/1 = 98g
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