An aqueous solution of mathrmHCl$\mathrm{HCl}$ with mathrmpH\ 1.0$\mathrm{pH}\ 1.0$ is diluted by adding equal volume of water (ignoring dissociation of water). The mathrmpH$\mathrm{pH}$ of mathrmHCl$\mathrm{HCl}$ solution would: (Given log 2 = 0.30$\log 2 = 0.30$)

### 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 HX_(aq) rightleftharpoons H^+_(aq) + X^-_(aq)$$HX_{(aq)} \rightleftharpoons H^{+}_{(aq)} + X^{-}_{(aq)}$$ K_a = frac[H^+][X^-][HX] = frac(Calpha)^2C(1-alpha) approx Calpha^2$$K_a = \frac{[H^+][X^-]}{[HX]} = \frac{(C\alpha)^2}{C(1-\alpha)} \approx C\alpha^2$$ ### Core Logic Official Answer Path Analysis: When the solution is diluted until mathrmpH = 6$\mathrm{pH} = 6$, the hydronium ion concentration becomes: [H^+] = 10^-6mathrm~M = C_textnewalpha_textnew$$[H^+] = 10^{-6}\mathrm{~M} = C_{\text{new}}\alpha_{\text{new}}$$ Applying the equilibrium constant expression without approximations for very high dilutions: K_a = fracCalpha^21-alpha = frac[H^+]alpha1-alpha = 4 times 10^-10$$K_a = \frac{C\alpha^2}{1-\alpha} = \frac{[H^+]\alpha}{1-\alpha} = 4 \times 10^{-10}$$ frac10^-6 cdot alpha1-alpha = 4 times 10^-10 implies 10^4 alpha = 4(1-alpha)$$\frac{10^{-6} \cdot \alpha}{1-\alpha} = 4 \times 10^{-10} \implies 10^4 \alpha = 4(1-\alpha)$$ 2500alpha = 1 - alpha implies 2501alpha = 1 implies alpha approx frac12500$$2500\alpha = 1 - \alpha \implies 2501\alpha = 1 \implies \alpha \approx \frac{1}{2500}$$ Now, substitute alpha$\alpha$ back to isolate the absolute concentration parameter C_textnew$C_{\text{new}}$: C_textnewalpha = 10^-6 implies C_textnew cdot left(frac12500right) = 10^-6$$C_{\text{new}}\alpha = 10^{-6} \implies C_{\text{new}} \cdot \left(\frac{1}{2500}\right) = 10^{-6}$$ C_textnew = 2500 times 10^-6 = 25 times 10^-4mathrm~M$$C_{\text{new}} = 2500 \times 10^{-6} = 25 \times 10^{-4}\mathrm{~M}$$ Therefore, comparing with x times 10^-4mathrm~M$x \times 10^{-4}\mathrm{~M}$ yields x = 25$x = 25$. ### Pattern Recognition When dealing with extreme dilution states where alpha$\alpha$ becomes large, you must avoid the standard (1-alpha) approx 1$(1-\alpha) \approx 1$ simplification step to ensure mathematically accurate answers. ### 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

### Core Logic Let's track the concentration changes for the reversible reaction starting with reactants mathrmH_2$\mathrm{H}_{2}$ and mathrmI_2$\mathrm{I}_{2}$: 1. As the forward reaction proceeds, the concentrations of reactants (mathrmH_2$\mathrm{H}_2$ and mathrmI_2$\mathrm{I}_2$) decrease over time. 2. Simultaneously, the concentration of the product (mathrmHI$\mathrm{HI}$) increases from zero. 3. Once dynamic equilibrium is attained, the rates of the forward and reverse reactions become equal. Consequently, the concentrations of all reactants and products become constant over time, appearing as horizontal lines on a concentration vs. time plot. Graph (2) correctly depicts the concentration of reactants smoothly decreasing and the product concentration increasing until they all plateau horizontally at dynamic equilibrium. ### Pattern Recognition On a concentration vs. time graph, look for lines that become perfectly horizontal after a certain point. This horizontal plateau signifies that equilibrium has been established. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

### Related Formula K_sp = x^x y^y s^x+y$$K_{sp} = x^x y^y s^{x+y}$$ ### Core Logic The dissolution equilibrium for chromium hydroxide is written as: Cr(OH)_3(s) rightleftharpoons Cr^3+(aq) + 3OH^-(aq)$$Cr(OH)_{3(s)} \rightleftharpoons Cr^{3+}(aq) + 3OH^{-}(aq)$$ If the molar solubility is denoted by s$s$: [Cr^3+] = s quad textand quad [OH^-] = 3s$$[Cr^{3+}] = s \quad \text{and} \quad [OH^{-}] = 3s$$ Substituting values into the expressions: K_sp = (s) cdot (3s)^3 = 27s^4$$K_{sp} = (s) \cdot (3s)^3 = 27s^4$$ Given K_sp = 1.6 times 10^-30$$K_{sp} = 1.6 \times 10^{-30}$$: 27s^4 = 1.6 times 10^-30$$27s^4 = 1.6 \times 10^{-30}$$ s = left( frac1.6 times 10^-3027 right)^1/4$$s = \left( \frac{1.6 \times 10^{-30}}{27} \right)^{1/4}$$ ### Pattern Recognition For a binary-quaternary salt of type AB_3$AB_3$, the relationship simplifies strictly to K_sp = 27s^4$K_{sp} = 27s^4$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium