Question

What are the final hydrogen ion concentration and pH of a solution obtained by mixing 400mL of 0.2M NaOH with 150mL of 0.1M H3PO4?

pKa's are 2.12, 7.21,12.32.

Answer 1

pH = 12.80

[H+] = 1.58 * 10^-13 M

Step-by-step explanation:

Step 1: Data given

Volume of 0.2M NaOH = 400 mL

Volume of 0.1M H3PO4 = 150 mL

Step 2: The balanced equation

H3PO4 + 3NaOH → Na3PO4 + 3H2O

For 1 mol H3PO4 we need 3 mol of NaOH to produce 1 mol Na3PO4 and 3 mol H2O

Step 3: Calculate moles H3PO4

Moles H3PO4 = molarity * volume

Moles H3PO4 = 0.1 M * 0.150 L

Moles H3PO4 = 0.015 moles

Step 4: Calculate moles NaOH

Moles NaOH = 0.2M * 0.400 L

Moles NaOH = 0.08 moles

For 1 mol H3PO4 we need 3 mol of NaOH to produce 1 mol Na3PO4 and 3 mol H2O

0.015 mol H3PO4 will react with 0.045 mol NaOH

Step 5: Calculate moles remaining

H3PO4 will be completely consumed

There will remain 0.08 - 0.045 = 0.035 moles of NaOH

Step 6: Calculate total volume

Total volume = 400 mL + 150 mL = 550 mL = 0.550 L

Step 7: Calculate molarity of the solution

Molarity = moles / volume

Molarity = 0.035 moles / 0.550 L

Molarity = 0.0636 M NaOH

Step 8: Calculate pOH

[OH-] = 0.0636M

pOH = -log [OH-]

pOH = -log(0.0636)

pOH= 1.20

Step 9: Calculate pH

pH = 14.00- pOH

pH = 14.00 - 1.20

pH = 12.80

[H+] = 10^-12.80

[H+] = 1.58 * 10^-13 M

Answer 2

Step-by-step explanation:

The chemical reaction equation will be as follows.

`H_(3)PO_(4) + 3NaOH \rightarrow Na_(3)PO_(4) + 3H_(2)O`

In this reaction, 1 mole of
`H_(3)PO_(4)` reacts with 3 mole NaOH. So, the number of moles of
`H_(3)PO_(4)` present in 150 ml of 0.1 M solution is calculated as follows.

No. of moles =
`(150)/(1000 * 0.1)`

= 0.015 mol

As it reacts with 3 moles of NaOH. Hence, no.. of moles of NaOH are:

`3 * 0.015 mol`

= 0.045 mol

So, moles of NaOH in 400 of 0.2 M NaH is as follows.

No. of moles =
`(400)/(1000 * 2)`

= 0.080 mol

Hence, no. of moles remained after the reaction are as follows.

(0.080 - 0.045) mol

= 0.035 mol NaOH in 550 ml (400 ml + 150 ml)

As molarity is the no. of moles present in liter of solution. Hence, molarity of NaOH is as follows.

Molarity =
`\frac{\text{no. of moles}}{\text{volume in liter}}`

=
`(0.035)/(550)`

= 0.0636 M

As,
`[OH^(-)]` = 0.0636 M. Hence, pOH will be 1.20.

As, pH + pOH = 14

pH = 14 - pOH

= 14 - 1.20

= 12.80

Also,
`[H^(+)] = 10^(-pH)`

So,
`[H^(+)] = 10^(-12.80)`

=
`1.58 * 10^(-13)` M

Thus, we can conclude that pH of the given solution is 12.80 and its hydrogen ion concentration is
`1.58 * 10^(-13)` M.