What is pH and how can we measure pH? Assuming complete ionization, calculate: t

Question

What is pH and how can we measure pH? Assuming complete ionization, calculate: the pH of 0.0001M HCI the pOH of 0.0001 M KOH the pH of 0.0001 M KOH the pH of 0.1M Ca(OH)_2 the pOH of 0.1 M 0.1M Ca(OH)_2 the pH of 0.1 M 0.1M Ca(OH)_2 Based on your calculations mark, which solution (listed above) will be acidic and, which one basic? You want to dilute 50 ml of 3.50 M H_2 SO_4 to get 2.00 M H_2 SO_4. Find the following: total volume of diluted solution; and what volume of water must be added to obtain correct dilution, what will be pH of diluted solution Find two applications where pH of water/wastewater has to be measured and known. You may use your book or internet resources. Write one paragraph about each application and state what pH is required/recommended for this particular application, and why keeping this pH is so important

Explanation / Answer

Solved first two problems, post multiple question to get the remaining answers

1) pH represents the acidity or basicity of the solution in logarithm scale, the pH value of 7 is neutral, the pH less than 7 is termed as acidic and pH greater than 7 is termed as basic

For calculating the pH

pH = -log[H+]

2)

a) pH of 0.0001M HCl

pH = -log[H+] = -log[0.0001] = 4

Hence the pH is equal to 4

b) pOH of 0.0001M KOH

pOH = -log[OH-] = -log[0.0001] = 4

Hence the pOH is equal to 4

c) pH of 0.0001M KOH

pOH = -log[OH-] = -log[0.0001] = 4

pH = 14 - pOH = 14 - 4 = 10

Hence the pH is equal to 10

d) Ca(OH)2-------Ca(2+) + 2OH-

pOH = -log[OH-] = -log[0.2] = 0.6989

pH = 14 - pOH = 14 - 0.6989 = 13.301

e) Ca(OH)2-------Ca(2+) + 2OH-

pOH = -log[OH-] = -log[0.2] = 0.6989

f) Ca(OH)2-------Ca(2+) + 2OH-

pOH = -log[OH-] = -log[0.2] = 0.6989

pH = 14 - pOH = 14 - 0.6989 = 13.301

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