### Related Formula textpH = -log_10[textH^+]$\text{pH} = -\log_{10}[\text{H}^+]$ ### Core Logic For the initial solution: textpH = 1.0 implies [textH^+]_1 = 10^-1 = 0.1 text M$\text{pH} = 1.0 \implies [\text{H}^+]_1 = 10^{-1} = 0.1 \text{ M}$ When we dilute the solution by adding an equal volume of water, the final volume is doubled (V_2 = 2V_1$V_2 = 2V_1$). Thus, the final concentration is halved: [textH^+]_2 = frac[textH^+]_12 = frac0.12 = 0.05 text M$[\text{H}^+]_2 = \frac{[\text{H}^+]_1}{2} = \frac{0.1}{2} = 0.05 \text{ M}$ Now, calculate the new textpH$\text{pH}$: textpH_2 = -log_10(0.05) = -log_10left(frac120right) = log_10 20 = log_10(10 times 2) = 1 + log_10 2$\text{pH}_2 = -\log_{10}(0.05) = -\log_{10}\left(\frac{1}{20}\right) = \log_{10} 20 = \log_{10}(10 \times 2) = 1 + \log_{10} 2$ textpH_2 = 1 + 0.30 = 1.30$\text{pH}_2 = 1 + 0.30 = 1.30$ ### Pattern Recognition Diluting any strong acid by 2$2$ times increases the textpH$\text{pH}$ by exactly log_10 2 approx 0.30$\log_{10} 2 \approx 0.30$. Thus, 1.0 + 0.3 = 1.3$1.0 + 0.3 = 1.3$ immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Ionic Equilibrium

### Related Formula For a weak acid solution: mathrmHA rightleftharpoons mathrmH^+ + mathrmA^-$\mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^-$ K_a = frac[mathrmH^+][mathrmA^-][mathrmHA]$K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}$ ### Step 1: Expressing Concentration Let the initial concentration be a$a$. At equilibrium: [mathrmHA] = a(1-x), quad [mathrmH^+] = ax, quad [mathrmA^-] = ax$[\mathrm{HA}] = a(1-x), \quad [\mathrm{H}^+] = ax, \quad [\mathrm{A}^-] = ax$ Substituting into the equilibrium expression: K_a = frac(ax)(x)1-x = [mathrmH^+] left(fracx1-xright)$K_a = \frac{(ax)(x)}{1-x} = [\mathrm{H}^+] \left(\frac{x}{1-x}\right)$ ### Step 2: Logarithmic Rearrangement Taking negative logarithms on both sides: -log K_a = -log [mathrmH^+] - logleft(fracx1-xright)$-\log K_a = -\log [\mathrm{H}^+] - \log\left(\frac{x}{1-x}\right)$ mathrmpK_a = mathrmpH - logleft(fracx1-xright)$\mathrm{pK}_a = \mathrm{pH} - \log\left(\frac{x}{1-x}\right)$ mathrmpH - mathrmpK_a = logleft(fracx1-xright)$\mathrm{pH} - \mathrm{pK}_a = \log\left(\frac{x}{1-x}\right)$ ### Pattern Recognition Sees: mathrmpH - mathrmpK_a$\mathrm{pH} - \mathrm{pK}_a$ for weak acid equilibrium. Shortcut: This is equivalent to the Henderson-Hasselbalch framework: mathrmpH = mathrmpK_a + logfrac[textSalt][textAcid] = mathrmpK_a + logfracx1-x$\mathrm{pH} = \mathrm{pK}_a + \log\frac{[\text{Salt}]}{[\text{Acid}]} = \mathrm{pK}_a + \log\frac{x}{1-x}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Ionic Equilibrium

### Related Formula textpH + textpOH = 14 implies textpOH = 14 - textpH$\text{pH} + \text{pOH} = 14 \implies \text{pOH} = 14 - \text{pH}$ [OH^-] = 10^-textpOH$[OH^-] = 10^{-\text{pOH}}$ textMoles of Mg(OH)_2 = frac[OH^-]2$\text{Moles of } Mg(OH)_2 = \frac{[OH^-]}{2}$ ### Core Logic 1. Convert the given pH into standard hydroxide concentration: textpOH = 14 - 10.0 = 4.0 implies [OH^-] = 10^-4 mathrm~mol cdot L^-1$\text{pOH} = 14 - 10.0 = 4.0 \implies [OH^-] = 10^{-4} \mathrm{~mol \cdot L^{-1}}$ 2. For 1.0 mathrm~L$1.0 \mathrm{~L}$ of solution, the number of moles of OH^-$OH^-$ required is 10^-4$10^{-4}$ moles. Since each formula unit of Mg(OH)_2$Mg(OH)_2$ releases 2 moles of OH^-$OH^-$ ions: textmoles of Mg(OH)_2 = frac10^-42 = 5 times 10^-5 text moles$\text{moles of } Mg(OH)_2 = \frac{10^{-4}}{2} = 5 \times 10^{-5} \text{ moles}$ 3. Convert this molar value into absolute mass units: textmass = 5 times 10^-5 times 58 mathrm~g = 2.9 times 10^-3 mathrm~g = 2.9 mathrm~mg$\text{mass} = 5 \times 10^{-5} \times 58 \mathrm{~g} = 2.9 \times 10^{-3} \mathrm{~g} = 2.9 \mathrm{~mg}$ Rounding to the nearest integer gives **3**. ### Pattern Recognition Always remember that Mg(OH)_2$Mg(OH)_2$ is a diacidic base. Forgetting to divide the hydroxide concentration by 2 is a frequent pitfall that leads to a value double the correct answer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Ionic Equilibrium

### Related Formula textpOH = textpK_textb + log frac[textSalt][textBase] $\text{pOH} = \text{p}K_{\text{b}} + \log \frac{[\text{Salt}]}{[\text{Base}]} $ textpH = 14 - textpOH $\text{pH} = 14 - \text{pOH} $ ### Core Logic Initially, the basic buffer solution contains: [textSalt] = [textNH4^+] = 0.10text mol, quad [textBase] = [textNH3] = 0.10text mol $[\text{Salt}] = [\text{NH}4^+] = 0.10\text{ mol}, \quad [\text{Base}] = [\text{NH}3] = 0.10\text{ mol} $ textpOHtextinitial = 4.745 + log frac0.100.10 = 4.745 $\text{pOH}{\text{initial}} = 4.745 + \log \frac{0.10}{0.10} = 4.745 $ When 0.05text mol$0.05\text{ mol}$ of strong acid textHCl$\text{HCl}$ is introduced, it reacts stoichiometrically with the weak base textNH_3$\text{NH}_3$: [cite: 1049, 1050] beginarrayrcccc & textNH3 & + & textH^+ & ightarrow & textNH4^+ \ textInitial (mol): & 0.10 & & 0.05 & & 0.10 \ textFinal (mol): & 0.05 & & 0 & & 0.15 endarray$\begin{array}{rcccc} & \text{NH}3 & + & \text{H}^+ & ightarrow & \text{NH}4^+ \ \text{Initial (mol):} & 0.10 & & 0.05 & & 0.10 \ \text{Final (mol):} & 0.05 & & 0 & & 0.15 \end{array}$ ### Step 1: Computing Post-Acid pOH and pH Recalculating via Henderson's equation: textpOHtextfinal = 4.745 + log frac0.150.05 = 4.745 + log 3 $\text{pOH}{\text{final}} = 4.745 + \log \frac{0.15}{0.05} = 4.745 + \log 3 $ The total shift value follows as: Delta textpOH = textpOHtextfinal - textpOHtextinitial = log 3 = 0.477 $\Delta \text{pOH} = \text{pOH}{\text{final}} - \text{pOH}{\text{initial}} = \log 3 = 0.477 $ Since textpH = 14 - textpOH$\text{pH} = 14 - \text{pOH}$: Delta textpH = -Delta textpOH = -0.477 $\Delta \text{pH} = -\Delta \text{pOH} = -0.477 $ Expressing the structural magnitude in scientific notation format: |Delta textpH| = 0.477 = 47.7 times 10^-2 approx 48 times 10^-2 $|\Delta \text{pH}| = 0.477 = 47.7 \times 10^{-2} \approx 48 \times 10^{-2} $ ### Pattern Recognition Buffer shifting rule: Adding an acid consumes base and builds salt. The base drops from 0.1$0.1$ to 0.05$0.05$ (halved), while salt grows from 0.1$0.1$ to 0.15$0.15$ (tripled). The ratio flips to 3, introducing a clean log 3$\log 3$ change factor into the solution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Ionic Equilibrium