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

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 = $\frac{150}{1000 \times 0.1}$  

                                      = 0.015 mol

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

                           $3 \times 0.015 mol$

                              = 0.045 mol

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

                No. of moles = $\frac{400}{1000 \times 2}$

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

                    = $\frac{0.035}{550}$        

                   = 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,          $[H^{+}] = 10^{-12.80}$    

                           = $1.58 \times 10^{-13}$ M

Thus, we can conclude that pH of the given solution is 12.80 and its hydrogen ion concentration is $1.58 \times 10^{-13}$ M.

Answer 2

Answer:

pH = 12.80  

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

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