you will prepare five tubes as follows: 4.0 mL of 0.10M NaH 2 PO 4 + 0 mL of 0.1

ID: 953239 • Letter: Y

Question

you will prepare five tubes as follows:

4.0 mL of 0.10M NaH2PO4 + 0 mL of 0.10M Na2HPO4

3.0 mL of 0.10M NaH2PO4 + 3.0 mL of 0.10M Na2HPO4

0 mL of 0.10M NaH2PO4 + 4.0 mL of 0.10M Na2HPO4

After adding some drops of indicator to each of the three solutions, you will add

2.0 mL 0.1 M of NaOH to tube 1

2.0 mL 0.1 M of HCl to tube 3.

Answer the following questions

TUBE 2:

What is the ratio of moles of H2PO4- / HPO42- in tube 2?

TUBE 1

How many moles of H2PO4- are initially in tube 1?

How many moles of NaOH were added to tube 1?

Write the net ionic equation showing the reaction between H2PO4- and NaOH.

How many moles of H2PO4- are left in tube 1 after addition of NaOH?

What is the ratio of moles of H2PO4- / HPO42- in tube 2 after addition of NaOH?

TUBE 3

How many moles of H2PO4- are initially in tube 3?

How many moles of HCl were added to tube 3?

Write the net ionic equation showing the reaction between HPO42- and HCl.

How many moles of HPO42- are left in tube 3 after addition of HCl?

What is the ratio of moles of H2PO4- / HPO42- in tube 3 after addition of HCl?

Tube 2

ratio of mols of H2PO4-/HPO4^2- = (0.1 x 3)/(0.1 x 3) = 1

Tube 1

initial moles of H2PO4- = 0.1 M x 4.0 ml = 0.4 mmol

moles of NaOH added = 0.1 M x 2 ml = 0.2 mmol

net ionic equation : H2PO4-(aq) + OH-(aq) ---> HPO4^2-(aq) + H2O(l)

mols of H2PO4- left after NaOH added = 0.2 mols

Final ratio of mols H2PO4-/HPO4^2- = 0.2/0.2 = 1

Tube 3

initial moles of H2PO4- = 0.1 x 0 = 0 mmol

moles of HCl added = 0.1 x 2 = 0.2 mmol

net ionic equation : HPO4^2-(aq) + H+(aq) <==> H2PO4-(aq)

moles of HPO4^2- left after HCl added = 0.2 mmol

ratio of mols of H2PO4-/HPO4^2- = 0.2/0.2 = 1

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