Can anyone help me solve for [H3O], [OH], pOH, and pH of solutions 1 and 2? 2. T

Question

Can anyone help me solve for [H3O], [OH], pOH, and pH of solutions 1 and 2?

2. Two solutions are prepared as follows: Solution 1: 250.0mL of 0.0800M H3PO4 Acid pk pk2 pk HP04 | 2.12 | 7.21 | 12.38 Solution 2: 2.015g of NaOH diluted to 750.0mL a. Complete the following table. Solution [H30] pOH 2 b. Write all three neutralization reactions of sodium hydroxide and phosphoric acid. Use stoichiometry and the Ka values provided to determine which equilibrium constant is most relevant when the solutions are mixed and calculate the final pH. c.

Explanation / Answer

Solution 1

Answer

[H3O+] = 0.02118M

[OH-] = 4.68×10-13M

[pH] = 1.67

[OH-] = 12.33

Explanation

First dissociation

H3PO4 <--------> H2PO4-   + H+

Ka1 = [H2PO4-][H+] /[H3PO4] = 7.59×10-3

at equillibrium

x2/(0.0800 - x) = 7.59×10-3

   x = 0.02114

Therefore,

[H3O+] = 0.02114M

[H2PO4-] = 0.02114M

Dissociation 2

H2PO42- <------> HPO4-   + H+

Ka2 = [HPO4-] [H+] / [H2PO4-] = 6.17×10-8

at equillibrium

x2/( 0.02114 - x) = 6.17×10-8

x =0.00003608

Therefore,

[H3O+] = 3.61×10-5M

[HPO42-] = 3.61×10-5M

third dissociation

HPO4- < - - - - - - > H+    + PO43-  

Ka3 = [H+] [PO43-] / [HPO4-] = 4.17×10-13

at equillibrium

x2/(3.61×10-5 - x) = 4.17×10-13

x = 3.88×10-9

Therefore,

[H3O+] = 3.88×10-9M

[PO42-] = 3.88×10-9M

Total [H3O+]= 0.02114M + 3.61×10-5M + 3.88×10-9M = 0.02118M

pH = log[H3O+] = - log(0.02118) = 1.67

pOH = 14 - pH = 14 - 1.67 = 12.33

[OH-] = - log[OH-]

- log[OH-] = 12.33

[OH-] =4.68×10-13M

Solution2

Answer

[H3O+] = 1.48×10-13M

[OH-] = 0.0672M

pH = 12.83

pOH = 1.17

Explanation

Mass of NaOH = 2.015g

No of moles of NaOH = 2.015g/40g/mol = 0.050375

Molarity of NaOH = (0.050375mol/750ml)×1000ml =0.0672M

Molarity of OH- = 0.0672M

pOH = - log[OH-] = - log(0.0672) = 1.17

pH = 14 - 1.17 = 12.83

-log[H3O+] = 12.83

[H3O+] = 1.48×10-13M

  

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