GATE Life Sciences Chemical Equilibria Syllabus

Give the formula for freezing point depression in terms of molality.

$$\Delta T_f = i\,K_f\,m$$ where $K_f$ is the cryoscopic (molal freezing point depression) constant, $m$ the molality, and $i$ the van't Hoff factor.

What is the cryoscopic constant $K_f$, and its thermodynamic expression?

$K_f$ is the molal depression constant: the freezing point depression of a $1\ \text{molal}$ ideal solution (units $\text{K}\cdot\text{kg}\cdot\text{mol}^{-1}$). $$K_f = \frac{R\,M_1\,T_f^{2}}{1000\,\Delta H_{fus}}$$

How is molar mass determined from freezing point depression?

$$M_2 = \frac{1000\,K_f\,w_2}{\Delta T_f\,w_1}$$ where $w_2$ is solute mass, $w_1$ solvent mass (g), and $\Delta T_f$ the observed depression.

Why is spreading salt ($\ce{NaCl}$ or $\ce{CaCl2}$) effective for de-icing roads?

The dissolved salt depresses the freezing point of water ($\Delta T_f = i K_f m$); with $i > 1$ for ionic salts the effect is large, so ice melts at temperatures below $0\,^\circ\text{C}$.

For the same molality, why is freezing point depression typically larger in magnitude per degree than boiling point elevation for the same solvent?

Because the cryoscopic constant is generally larger than the ebullioscopic constant ($K_f > K_b$) for most solvents (e.g., water: $K_f = 1.86$, $K_b = 0.52\ \text{K kg mol}^{-1}$), owing to the smaller enthalpy of fusion relative to vaporization.

Define the solubility product constant $K_{sp}$.

For a sparingly soluble salt $\ce{A_xB_y(s) <=> xA^{y+} + yB^{x-}}$, the solubility product is $$K_{sp} = [\text{A}^{y+}]^{x}\,[\text{B}^{x-}]^{y}$$ the equilibrium product of ionic concentrations in a saturated solution, each raised to its stoichiometric coefficient.

Write $K_{sp}$ in terms of molar solubility $s$ for $\ce{AB}$, $\ce{AB2}$, and $\ce{A2B3}$ type salts.

$\ce{AB}$: $K_{sp} = s^{2}$. $\ce{AB2}$ (or $\ce{A2B}$): $K_{sp} = 4s^{3}$. $\ce{A2B3}$: $K_{sp} = 108\,s^{5}$.

Derive the molar solubility of $\ce{Ag2CrO4}$ in terms of $K_{sp}$.

$\ce{Ag2CrO4 <=> 2Ag+ + CrO4^{2-}}$, so $K_{sp} = (2s)^{2}(s) = 4s^{3}$, giving $$s = \sqrt[3]{\frac{K_{sp}}{4}}$$

What is the ionic product $Q_{sp}$, and how does comparing it with $K_{sp}$ predict precipitation?

$Q_{sp}$ (reaction quotient) is computed from actual ion concentrations the same way as $K_{sp}$. If $Q_{sp} > K_{sp}$: precipitation occurs (supersaturated). If $Q_{sp} = K_{sp}$: saturated equilibrium. If $Q_{sp} < K_{sp}$: unsaturated, no precipitate (more solid can dissolve).

Distinguish solubility ($s$) from solubility product ($K_{sp}$).

Solubility $s$ is the amount of salt that dissolves per unit volume to make a saturated solution (depends on common ions, temperature, pH). $K_{sp}$ is the equilibrium constant for dissolution and is constant at a given temperature, independent of added common ions.

Why does $K_{sp}$ generally increase with temperature for most salts?

Dissolution of most salts is endothermic; by Le Chatelier's principle, raising the temperature shifts the dissolution equilibrium forward, increasing solubility and hence $K_{sp}$. (Exceptions exist for salts whose dissolution is exothermic.)

State the common ion effect.

The common ion effect is the suppression of the dissociation (or solubility) of a weak electrolyte or sparingly soluble salt upon adding a strong electrolyte that provides an ion common to the equilibrium, shifting it backward (Le Chatelier).

How does adding $\ce{NaCl}$ affect the solubility of $\ce{AgCl}$, and why?

It decreases the solubility of $\ce{AgCl}$. The added $\ce{Cl-}$ (common ion) increases $[\ce{Cl-}]$, so to keep $K_{sp}=[\ce{Ag+}][\ce{Cl-}]$ constant, $[\ce{Ag+}]$ must fall, shifting $\ce{AgCl(s) <=> Ag+ + Cl-}$ backward and precipitating more $\ce{AgCl}$.

Give the quantitative effect: solubility $s$ of $\ce{AgCl}$ in a solution already $C$ molar in $\ce{Cl-}$.

With the common ion, $[\ce{Ag+}] = s$ and $[\ce{Cl-}] \approx C$, so $K_{sp} = s\cdot C$ and $$s = \frac{K_{sp}}{C}$$ much smaller than $\sqrt{K_{sp}}$ in pure water.

How is the common ion effect used in the qualitative analysis of cations (group separation)?

Adding a common ion (e.g., $\ce{H2S}$ in presence of $\ce{HCl}$, or $\ce{NH4Cl}$ before $\ce{NH4OH}$) lowers the concentration of $\ce{S^{2-}}$ or $\ce{OH-}$, controlling which sulfides/hydroxides exceed their $K_{sp}$ and precipitate, allowing selective separation of cation groups.

Define salt hydrolysis.

Salt hydrolysis is the reaction of the cation and/or anion of a dissolved salt with water to produce $\ce{H3O+}$ or $\ce{OH-}$ ions, making the solution acidic, basic, or neutral; it is essentially the reverse of neutralization.

Classify the four types of salts by the strength of their parent acid and base, with the resulting solution pH.

Strong acid + strong base (e.g., $\ce{NaCl}$): neutral, $pH = 7$. Strong acid + weak base (e.g., $\ce{NH4Cl}$): acidic, $pH < 7$. Weak acid + strong base (e.g., $\ce{CH3COONa}$): basic, $pH > 7$. Weak acid + weak base (e.g., $\ce{CH3COONH4}$): depends on relative $K_a$ and $K_b$.

Why is a solution of sodium acetate ($\ce{CH3COONa}$) basic? Write the hydrolysis equation.

The acetate ion (conjugate base of a weak acid) hydrolyzes: $$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$ producing excess $\ce{OH-}$, making the solution basic ($pH > 7$). The $\ce{Na+}$ ion does not hydrolyze.

Define the hydrolysis constant $K_h$ and relate it to $K_w$, $K_a$, $K_b$ for a salt of a weak acid and strong base.

$$K_h = \frac{K_w}{K_a}$$ where $K_w$ is the ionic product of water and $K_a$ the dissociation constant of the weak acid. The larger $K_h$, the greater the extent of hydrolysis.

For a salt of weak acid + strong base, give the degree of hydrolysis $h$ and the pH in terms of $K_a$, $K_w$, and concentration $C$.

$$h = \sqrt{\frac{K_w}{K_a\,C}}, \qquad pH = \frac{1}{2}\left(pK_w + pK_a + \log C\right)$$

For a salt of strong acid + weak base, give the degree of hydrolysis $h$ and the pH in terms of $K_b$, $K_w$, and concentration $C$.

$$h = \sqrt{\frac{K_w}{K_b\,C}}, \qquad pH = \frac{1}{2}\left(pK_w - pK_b - \log C\right)$$

For a salt of weak acid + weak base, give $K_h$, the degree of hydrolysis, and the pH.

$$K_h = \frac{K_w}{K_a K_b}, \qquad h = \sqrt{\frac{K_w}{K_a K_b}} \ (\text{≈ independent of } C), \qquad pH = \frac{1}{2}\left(pK_w + pK_a - pK_b\right)$$

Define pH and pOH.

$$pH = -\log_{10}[\ce{H3O+}], \qquad pOH = -\log_{10}[\ce{OH-}]$$ pH is the negative base-10 logarithm of the molar hydronium ion concentration.

State the relationship between pH, pOH, and $pK_w$ at $25\,^\circ\text{C}$.

$$pH + pOH = pK_w = 14$$ since $K_w = [\ce{H3O+}][\ce{OH-}] = 1.0\times 10^{-14}\ \text{at } 25\,^\circ\text{C}$.

Why does the pH of pure water decrease below 7 as temperature rises, even though it stays neutral?

Water's self-ionization is endothermic, so $K_w$ increases with temperature; $[\ce{H3O+}]$ and $[\ce{OH-}]$ both rise (staying equal, so still neutral), but the larger $[\ce{H3O+}]$ gives a pH below 7.

Give the pH of a weak monoprotic acid solution of concentration $C$ and dissociation constant $K_a$ (small dissociation).

$[\ce{H+}] = \sqrt{K_a C}$, so $$pH = \frac{1}{2}\left(pK_a - \log C\right) = \frac{1}{2}pK_a - \frac{1}{2}\log C$$

Define a buffer solution and name the two common types.

A buffer is a solution that resists changes in pH on adding small amounts of acid or base. The two types are: (1) acidic buffer (weak acid + its salt with strong base, e.g., $\ce{CH3COOH/CH3COONa}$), and (2) basic buffer (weak base + its salt with strong acid, e.g., $\ce{NH4OH/NH4Cl}$).

State the Henderson–Hasselbalch equation for an acidic buffer.

$$pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]} = pK_a + \log\frac{[\text{conjugate base}]}{[\text{acid}]}$$

State the Henderson–Hasselbalch equation for a basic buffer and the condition for maximum buffer capacity.

$$pOH = pK_b + \log\frac{[\text{salt}]}{[\text{base}]}$$ Maximum buffer capacity occurs when $[\text{salt}] = [\text{base}]$ (equal components), where $pH = pK_a$ and the buffer best resists pH change.

Explain how an acetic acid / acetate buffer neutralizes added strong acid and added strong base.

Added $\ce{H+}$ is consumed by the conjugate base: $\ce{CH3COO- + H+ -> CH3COOH}$. Added $\ce{OH-}$ is consumed by the weak acid: $\ce{CH3COOH + OH- -> CH3COO- + H2O}$. In both cases the ratio [salt]/[acid] changes only slightly, so pH stays nearly constant.

List important applications of buffers.

Maintaining blood pH near 7.4 (carbonic acid–bicarbonate buffer), enzymatic and biochemical reactions, fermentation, electroplating baths, agriculture/soil pH control, manufacture of dyes and pharmaceuticals, calibration of pH meters, and analytical/qualitative chemistry.

Which buffer system maintains human blood pH, and what is its useful pH range relative to $pK_a$?

The carbonic acid–bicarbonate buffer, $\ce{H2CO3/HCO3-}$ (with $pK_{a1} \approx 6.1$), maintains blood pH $\approx 7.4$. A buffer is generally effective within $pH = pK_a \pm 1$.