Calculate the pH of equilibrium aqueous solutions of the following:
a) 10-4 M sodium hydroxide, NaOH
b) 1.0 mg/L sulfuric acid (H2S04)
Note: NaOH is a strong base and H2SO4 is a strong acid. Both undergo complete dissociation
when added to water according to the following chemical equations: NaOH <--> Na+ + OH-
H2SO4 <--> 2 H+ + SO4-2
Solution
a) pH of NaOH
pOH = -log [ 10-4]
pOH = 4
pH + pOH = 14
pH = 14-4 = 10
b) 1 mg/L = 0.001 mg/L
0.001/ 98.1 = 1.02 x 10-5 M
H+ = 2 x 1.02 x 10-5 M = 2.04 x 10-5 M
pH = -log[2.04 x 10-5 ]
pH = 4.69.
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Calculate the pH of equilibrium aqueous solutions of the following.pdf
1. Calculate the pH of equilibrium aqueous solutions of the following:
a) 10-4 M sodium hydroxide, NaOH
b) 1.0 mg/L sulfuric acid (H2S04)
Note: NaOH is a strong base and H2SO4 is a strong acid. Both undergo complete dissociation
when added to water according to the following chemical equations: NaOH <--> Na+ + OH-
H2SO4 <--> 2 H+ + SO4-2
Solution
a) pH of NaOH
pOH = -log [ 10-4]
pOH = 4
pH + pOH = 14
pH = 14-4 = 10
b) 1 mg/L = 0.001 mg/L
0.001/ 98.1 = 1.02 x 10-5 M
H+ = 2 x 1.02 x 10-5 M = 2.04 x 10-5 M
pH = -log[2.04 x 10-5 ]
pH = 4.69
![Calculate the pH of equilibrium aqueous solutions of the following:
a) 10-4 M sodium hydroxide, NaOH
b) 1.0 mg/L sulfuric acid (H2S04)
Note: NaOH is a strong base and H2SO4 is a strong acid. Both undergo complete dissociation
when added to water according to the following chemical equations: NaOH <--> Na+ + OH-
H2SO4 <--> 2 H+ + SO4-2
Solution
a) pH of NaOH
pOH = -log [ 10-4]
pOH = 4
pH + pOH = 14
pH = 14-4 = 10
b) 1 mg/L = 0.001 mg/L
0.001/ 98.1 = 1.02 x 10-5 M
H+ = 2 x 1.02 x 10-5 M = 2.04 x 10-5 M
pH = -log[2.04 x 10-5 ]
pH = 4.69](https://image.slidesharecdn.com/calculatethephofequilibriumaqueoussolutionsofthefollowing-230326050615-cad3ff0e/85/calculate-the-ph-of-equilibrium-aqueous-solutions-of-the-followingpdf-1-320.jpg)
