Using the formula for weak acids, pH = 0.5(pKa - log[C]), where pKa = -log(1.8 x 10^-5) ≈ 4.74. Thus, pH = 0.5(4.74 - log(0.1)) = 3.87.

Using Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]); pKa of acetic acid = 4.76, so pH = 4.76 + log(1) = 4.76

Using the formula for weak acids, pH = 0.5(pKa - logC) = 0.5(4.74 - log(0.1)) = 3.87.

Strong acid and strong base neutralize each other, resulting in a neutral solution with pH = 7.

Using the formula Ka = [H+]^2 / [HA], where [H+] = 10^(-4) M and [HA] = 0.1 M, we find Ka = (10^-4)^2 / 0.1 = 1 x 10^-5.

HCl and NaOH neutralize each other, resulting in a neutral solution with pH = 7.

Using the formula for weak bases, pH = 14 - pKb = 14 - (14 - 4.75) = 5.75.

[H+] = 10^(-pH) = 10^(-3) = 0.001 M

Adding a common ion decreases the solubility of a salt in a saturated solution due to the common ion effect.

A buffer solution is designed to resist changes in pH, so adding a strong acid will cause only a small change in pH.

Adding a strong base to a buffer solution will increase the pH, but the buffer will resist significant changes in pH due to the presence of the weak acid.

Diluting a strong acid decreases its concentration, which increases the pH (making it less acidic).

Dilution of a strong acid decreases its concentration, thus increasing the pH.

Dilution of a weak acid decreases its concentration, which shifts the equilibrium to the right, increasing the concentration of H+ ions and thus increasing the pH.

Kb = Kw / Ka = 1.0 x 10^-14 / 5.6 x 10^-10 = 1.8 x 10^-5

The solubility product constant (Ksp) for Ag2SO4 is given by Ksp = [Ag+]^2[SO4^2-] because the dissociation is Ag2SO4 ⇌ 2Ag+ + SO4^2-.

Ksp = [Ag+][Cl-] = (1.0 x 10^-5)(1.0 x 10^-5) = 1.0 x 10^-10.

For NaOH, which is a strong base, pOH = -log[OH-]. [OH-] = 0.01 M, so pOH = 2. Therefore, pH = 14 - pOH = 14 - 2 = 12.

Sodium acetate is a salt of a weak acid (acetic acid) and a strong base (sodium hydroxide). The pH can be calculated using the formula pH = 7 + 0.5(pKa - log[C]), where pKa of acetic acid is 4.76.

pH = 14 - pOH; pOH = -log[OH-] = -log(0.01) = 2; pH = 14 - 2 = 12.

H2SO4 is a strong acid and dissociates completely, so pH = -log(0.01) = 2.

H2SO4 is a strong acid and dissociates completely; [H+] = 0.05 M, so pH = -log(0.05) ≈ 1.3

pOH = -log[OH-] = -log(0.1) = 1; pH = 14 - pOH = 14 - 1 = 13

Using the formula for weak acids, pH = 0.5(pKa - log[C]) = 0.5(4.74 - log(0.1)) = 4.76.

NH4Cl is a salt of a weak base and a strong acid; it hydrolyzes to give H+, resulting in a pH < 7.

The pH of a strong acid like HCl is calculated using the formula pH = -log[H+]. For 0.1 M HCl, [H+] = 0.1 M, so pH = -log(0.1) = 1.

KHP is a weak acid; its pKa is approximately 5.0, so the pH of a 0.1 M solution is around 5.0.

pH = 7 + 0.5(pKa + log[C]) = 7 + 0.5(4.76 + log(0.1)) = 9.24

NaHCO3 is a weak base; its pH is around 8.4.

Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]). Here, pKa ≈ 4.74, so pH = 4.74 + log(1) = 4.74.