Concentration units, acids and bases, and pH: VCE Chemistry Unit 2

What this dot point is asking

VCAA wants you to express the concentration of a solution in any of the standard VCE units, perform dilution calculations, describe the Brønsted-Lowry model of acids and bases including conjugate acid-base pairs, distinguish strong vs weak and concentrated vs dilute, and calculate pH from a hydrogen-ion concentration.

The answer

Concentration units

Pick the unit that matches the question's data:

Conversions you will need:

• mol L^-1 to g L^-1: multiply by molar mass M.
• ppm to mg/L for dilute aqueous solutions: 1 ppm = 1 mg/L.
• %m/v to g L^-1: 1 %m/v = 10 g L^-1.

Dilution

Adding more solvent decreases concentration but not the moles of solute. The dilution formula follows from n = cV being constant:

c1 V1 = c2 V2

Use it directly when a more concentrated stock is diluted to a less concentrated working solution. Units of V can be anything (mL or L) as long as you use the same units on both sides.

The Brønsted-Lowry model

A Brønsted-Lowry acid is a proton donor; a Brønsted-Lowry base is a proton acceptor. An acid-base reaction is a proton transfer:

HA + B <=> A^- + HB^+

The pair (HA, A^-) is a conjugate acid-base pair: HA is the acid, A^- is its conjugate base. Similarly (HB^+, B) is a pair: HB^+ is the conjugate acid of base B.

Common examples:

Water is amphiprotic: it can act as either an acid (donating H+ to give OH^-) or a base (accepting H+ to give H3O^+).

Strong vs weak; concentrated vs dilute

Strong = ionises (or dissociates) completely in water. All HCl molecules become H+ and Cl^-. The position of the ionisation equilibrium lies essentially fully to the right.

Weak = ionises partially. Only a small fraction of CH3COOH molecules give up a proton at any moment; the rest stay molecular.

Concentrated = a high number of moles of solute per litre of solution.

Dilute = a low number of moles of solute per litre.

These are independent axes. A weak acid can be concentrated (concentrated acetic acid, "glacial" acetic acid, is 17 mol L^-1 and weak); a strong acid can be dilute (0.0001 mol L^-1 HCl is dilute and strong).

Common strong acids you should recognise: HCl, HBr, HI, HNO3, H2SO4 (first proton). Common strong bases: NaOH, KOH, Ca(OH)2 (where dissolved), Ba(OH)2.

pH

pH is defined as:

pH = -log10([H+])

where [H+] is in mol L^-1. For pure water at 25 deg C, [H+] = 10^-7 mol L^-1 and pH = 7.

Useful shortcuts:

• A pH of 1 has [H+] = 0.1; pH of 2 has [H+] = 0.01; each pH unit is a factor of 10 in [H+].
• For a strong monoprotic acid, [H+] equals the acid concentration. 0.05 mol L^-1 HCl gives [H+] = 0.05, pH = 1.30.
• For a strong base, calculate [OH-] then use [H+][OH-] = 10^-14 at 25 deg C, then pH = -log[H+]. Or pOH = -log[OH-] and pH = 14 - pOH.
• For a weak acid or base, you cannot assume [H+] equals concentration. Weak acids have higher pH than strong acids of the same concentration; weak bases have lower pH than strong bases of the same concentration.

Examples in context

Example 1. Yarra River pH monitoring at Dights Falls. Melbourne Water posts continuous water-quality data from Dights Falls and other Yarra River sites. After heavy rain, urban stormwater runoff can carry sufficient $\text{CO}_2$ and humic acids to drop the pH from a baseline near $7.2$ to as low as $5.8$ within hours. A pH of $5.8$ corresponds to $[\text{H}^+] = 10^{-5.8} = 1.6 \times 10^{-6} \, \text{mol/L}$, about $25$ times higher than at pH $7.2$. The ANZECC water-quality guideline for protection of $95\%$ of freshwater species is pH $6.5$ to $8.0$. The probe is calibrated daily with pH $4.0$ phthalate and pH $7.0$ phosphate buffers traceable to the National Measurement Institute in Lindfield.

Example 2. Industrial concentration of hydrochloric acid at Orica Yarraville. Orica's Yarraville site supplies concentrated hydrochloric acid (around $32\%$ m/m, density $1.16 \, \text{g/mL}$) used for steel pickling. Calculating molarity: in $1 \, \text{L}$ there is $1160 \, \text{g}$ of solution, of which $371 \, \text{g}$ is $\text{HCl}$, giving $n = 371 / 36.5 = 10.2$ mol. So $c = 10.2 \, \text{mol/L}$, which is the standard "concentrated HCl" used in school labs after dilution. To prepare $500 \, \text{mL}$ of $1.00 \, \text{mol/L}$ working solution from this stock requires $V_1 = c_2 V_2 / c_1 = (1.00 \times 0.500) / 10.2 = 0.0490 \, \text{L}$ or $49.0 \, \text{mL}$ of concentrate added to water with gentle stirring.

Try this

Q1. Calculate the pH of a solution prepared by diluting $25.0 \, \text{mL}$ of $0.100 \, \text{mol/L}$ HCl to $250 \, \text{mL}$ with water. [2 marks]

• Cue. $c_2 = 0.100 \times 25.0 / 250 = 0.0100 \, \text{mol/L}$; pH $= - \log (0.0100) = 2.00$.

Q2. A water sample contains $5 \, \text{ppm}$ of calcium ions. (a) Express this in $\text{mg/L}$ and $\text{mol/L}$. (b) Explain why $1 \, \text{ppm}$ approximates $1 \, \text{mg/L}$ for dilute aqueous solutions. [3 marks]

• Cue. (a) $5 \, \text{mg/L}$; $c = 5 \times 10^{-3} / 40.1 = 1.25 \times 10^{-4} \, \text{mol/L}$. (b) Density of water close to $1 \, \text{g/mL}$, so $1 \, \text{mg}$ per kg $\approx 1 \, \text{mg}$ per L.

Q3. A swimming pool contains $50{,}000 \, \text{L}$ of water at pH $8.4$. (a) Calculate $[\text{H}^+]$ and $[\text{OH}^-]$. (b) Determine moles of $\text{HCl}$ needed to lower the pH to $7.4$. (c) Identify the conjugate base of $\text{HCl}$. [2+2+2 marks]

• Cue. (a) $[\text{H}^+] = 4.0 \times 10^{-9}$, $[\text{OH}^-] = 2.5 \times 10^{-6} \, \text{mol/L}$. (b) Approximation: drop by $1$ pH unit means $[\text{H}^+]$ increases ten-fold; for the buffer-free water, add roughly $0.5$ mol of HCl into $50{,}000 \, \text{L}$ as a rough estimate. (c) $\text{Cl}^-$.