1) What is the concentration of iron ions (in M) in a 5.16 M solution of Fe3(PO4
ID: 1001603 • Letter: 1
Question
1) What is the concentration of iron ions (in M) in a 5.16 M solution of Fe3(PO4)2?
2) Assuming volumes are additive, how much water (in mL) should be added to 31.7 mL of a 1.16 M solution of hydrochloric acid (HCl) to produce a 0.100 M solution?
3) Assuming volumes are additive, how much water (in mL) should be added to 13.6 mL of a 2.58 M solution of sodium hydroxide (NaOH) to produce a 0.250 M solution? Please give your answer with three significant digits even if it is not justified
4) What volume (in mL!!!) of 1.09 M NaOH is required to react with 23.2 mL of 1.25 M citric acid according to the following reaction?
H3C6H5O7 (aq) + 3NaOH (aq) 3H2O(l) + Na3C6H5O7 (aq)
5) What volume (in mL!!!) of 2.37 M HCl is required to react with 3.46 g of zinc (65.41 g/mol) according to the following reaction?
Zn(s) + 2 HCl (aq) ZnCl2(aq) + H2(g)
Explanation / Answer
1) What is the concentration of iron ions (in M) in a 5.16 M solution of Fe3(PO4)2?
Ans: The concentration of iron ion = 3 X concentration of salt as each mole will give three moles of iron ion
Concentration of ion = 3 X 5.16 = 15.48 M
2) Assuming volumes are additive, how much water (in mL) should be added to 31.7 mL of a 1.16 M solution of hydrochloric acid (HCl) to produce a 0.100 M solution?
Answer: The initial concentration of HCl = 1.16 M
Volume = 31.7 mL
Required concentration = 0.1M
Volume of water required = ?
M1V1 = M2V2
V2 = 1.16 X 31.7 / 0.1 = 367.72 mL
so volume of water required = 367.72-31.7 = 336.02mL
3) Assuming volumes are additive, how much water (in mL) should be added to 13.6 mL of a 2.58 M solution of sodium hydroxide (NaOH) to produce a 0.250 M solution? Please give your answer with three significant digits even if it is not justified
Answer: The same approach will be applied here
M1V1 = M2V2
V2 = 13.6 X 2.58 / 0.25 = 140.352 mL
So volume of water = 140.352 - 13.6 = 126.752 mL
4) What volume (in mL!!!) of 1.09 M NaOH is required to react with 23.2 mL of 1.25 M citric acid according to the following reaction?
H3C6H5O7 (aq) + 3NaOH (aq) 3H2O(l) + Na3C6H5O7 (aq)
Answer: As per the stochiometry of equation 3 moles of NaOH will react with one mole of citric acid
Moles of citric acid = Molarity X volume = 23.2 X 1.25 = 29 millimoles
so moles of NaOH = 3X29 = 87 millimoles
Concentration of NaOH = 1.09
so volume required = Moles / Molarity = 87 millimoles / 1.09 = 79.82 mL
5) What volume (in mL!!!) of 2.37 M HCl is required to react with 3.46 g of zinc (65.41 g/mol) according to the following reaction?
Zn(s) + 2 HCl (aq) ZnCl2(aq) + H2(g
one mole of Zn will react with 2 moles of HCl
Moles of Zn = Mass / Molecular weight = 3.46 / 65.41 = 0.0529
so moles of Hcl required =2 X 0.0529 = 0.1058 moles
Volume = Moles / Moalrity = 0.1058 / 2.37 = 0.0446 Litres = 44.6 mL
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