Solutions, Acids & Bases — Science

Molarity = moles of solute ÷ litres of solution, so convert grams to moles and millilitres to litres first.

Moles NaCl = 11.7 g ÷ 58.44 g/mol; volume = 250 mL = 0.250 L.

Divide the moles you get by 0.250 L.

0.8 mol/L

Moles of NaCl = 11.7 ÷ 58.44 = 0.200 mol. Volume = 250 mL = 0.250 L.

Molarity = 0.200 mol ÷ 0.250 L.

= 0.80 mol/L (0.80 M).

Diluting doesn't change the moles of acid, so use M₁V₁ = M₂V₂ and solve for the concentrated volume V₁.

V₁ = (M₂ × V₂) ÷ M₁ = (0.150 × 500.0) ÷ 12.0.

You can keep the volume in mL on both sides since the unit cancels; just compute (0.150 × 500.0) ÷ 12.0.

6.25 mL

Dilution conserves moles: M₁V₁ = M₂V₂, so V₁ = (M₂V₂) ÷ M₁.

V₁ = (0.150 M × 500.0 mL) ÷ 12.0 M = 75.0 ÷ 12.0.

= 6.25 mL of concentrated acid (then add water up to 500 mL).

pH is the negative base-10 logarithm of the hydrogen-ion concentration.

Because 2.5 × 10⁻³ lies between 10⁻³ and 10⁻² M, the pH must fall between 2 and 3.

Compute −log(2.5 × 10⁻³) = 3 − log(2.5), and log(2.5) ≈ 0.40.

2.6

pH = −log[H⁺] = −log(2.5 × 10⁻³).

−log(2.5 × 10⁻³) = 3 − log(2.5) = 3 − 0.398.

pH ≈ 2.60 (acidic, as expected for a value under 7).

At the neutralization point the moles of acid equal the moles of base (they react 1 : 1).

Moles of NaOH = molarity × litres = 0.100 M × 0.0300 L.

Set moles of HCl equal to that, then divide by the acid's volume in litres (0.0250 L).

0.12 mol/L

Moles of NaOH = 0.100 M × 0.0300 L = 0.00300 mol; the 1 : 1 ratio means 0.00300 mol of HCl was present.

Molarity of HCl = moles ÷ volume = 0.00300 mol ÷ 0.0250 L.

= 0.120 mol/L.

First find pOH from the hydroxide-ion concentration, then use pH + pOH = 14 to get pH.

pOH = −log(0.010) = −log(10⁻²).

Once pOH = 2, subtract it from 14.

12

Because NaOH ionizes completely, [OH⁻] = 0.010 M = 10⁻² M, so pOH = −log(10⁻²) = 2.

pH = 14 − pOH = 14 − 2.

pH = 12 (strongly basic, as expected for a strong base).

Solubility is a ratio — grams of solute per 100 g of water — so scale it up to the amount of water you actually have.

You have 250 g of water, which is 2.5 times the reference 100 g.

Maximum KCl = 34 g × (250 ÷ 100).

85 g

Solubility = 34 g KCl per 100 g water; 250 g of water is 250 ÷ 100 = 2.5 times as much.

Maximum dissolved = 34 g × 2.5.

= 85 g of KCl.

Lowest pH means most acidic — the largest hydrogen-ion concentration [H⁺].

A base and a neutral salt won't have a low pH; that leaves the two acids. Compare a strong acid (fully ionized) with a weak acid (only partly ionized) at the same concentration.

The strong acid releases nearly all of its 0.10 M as H⁺, while the weak acid releases only a small fraction — so the strong acid reaches the lowest pH.

0.10 M HCl (a strong acid)

NaOH is basic (high pH) and NaCl is neutral (pH ≈ 7), so neither can be the lowest.

Both remaining choices are acids, but HCl is strong and ionizes almost completely (≈ 0.10 M H⁺), while acetic acid is weak and ionizes only slightly (far less H⁺).

The most H⁺, and therefore the lowest pH, belongs to 0.10 M HCl (choice B).