1) The pH of a 0.03 M solution of nitric acid can be calculated using the formula:
pH = -log[H+]
where [H+] is the concentration of hydronium ions in the solution.
Since nitric acid is a strong acid, it completely dissociates in water to form H+ ions and NO3- ions. Therefore, the concentration of H+ ions in a 0.03 M solution of nitric acid is also 0.03 M.
Substituting this value into the formula, we get:
pH = -log(0.03) ≈ 1.52
Therefore, the pH of a 0.03 M solution of nitric acid is approximately 1.52.
2) The hydronium ion concentration of a sulfuric acid solution with a pH of 5.43 can be calculated using the formula:
pH = -log[H+]
where [H+] is the concentration of hydronium ions in the solution.
Substituting the given pH value into the formula, we get:
5.43 = -log[H+]
Taking the antilogarithm of both sides, we get:
[H+] = 10^(-5.43)
[H+] ≈ 2.15 x 10^(-6) M
Therefore, the hydronium ion concentration of the sulfuric acid solution is approximately 2.15 x 10^(-6) M.
3) The pOH of a 0.025 M solution of sodium hydroxide can be calculated using the formula:
pOH = -log[OH-]
where [OH-] is the concentration of hydroxide ions in the solution.
Since sodium hydroxide is a strong base, it completely dissociates in water to form Na+ ions and OH- ions. Therefore, the concentration of OH- ions in a 0.025 M solution of sodium hydroxide is also 0.025 M.
Substituting this value into the formula, we get:
pOH = -log(0.025) ≈ 1.60
Since pH + pOH = 14, we can calculate the pH of the solution as:
pH = 14 - pOH = 14 - 1.60 ≈ 12.40
Therefore, the pH of a 0.025 M solution of sodium hydroxide is approximately 12.40.
4) The pH of a 0.002 M solution of lithium hydroxide can be calculated using the formula:
pH = 14 - pOH
where pOH is the negative logarithm of the hydroxide ion concentration [OH-].
Since lithium hydroxide is a strong base, it completely dissociates in water to form Li+ ions and OH- ions. Therefore, the concentration of OH- ions in a 0.002 M solution of lithium hydroxide is also 0.002 M.
Substituting this value into the formula, we get:
pOH = -log(0.002) ≈ 2.70
Therefore, the pH of a 0.002 M solution of lithium hydroxide is approximately:
pH = 14 - pOH = 14 - 2.70 ≈ 11.30