HCOONa + HOH  ⇌ NaOH + H-COOH
                               strong base weak acid
Basic in nature.
C₆H₅NH₃Cl + HOH  ⇌ H₃O⁺ + C₆H₅ - NH₂ + Cl^-
                                      acidic
KCN + H - OH ⇌ KOH + HCN
Base                  strong base weak acid
basic, acidic, basic is correct.

Basic buffer is the solution which has weak base and its salt.
NH₄OH + HCI → NH₄CI +H₂O + NH₄OH
200 ml   100 ml    salt       100 ml  weak base

H₂O + H₂O ⇌  H₃O⁺ + OH^-
acid 1  base 1  acid 2   base 2
HF + H₂O ⇌  H₃O⁺ + F^-
acid 1  base 1  acid 2   base 2
∴ Conjugate bases are OH^- and F^- respectively.

Ca(OH)₂ ⇌ Ca²⁺ + 2OH^-
Given that pH = 9
pOH = 14 - 9 = 5
[pOH = - log₁₀ [OH]]
[OH^-] = 10 ^[pOH]
[OH] = 10^-5 M
Ksp = [Ca²⁺] [OH^-]
= 10^-5/2 x (10^-5)²  = 0.5 x 10^-15

\(\mathrm{Ag}_{2} \mathrm{C_2O_4} \rightleftharpoons 2 \mathrm{Ag}_{}^{+}+\mathrm{C_2O}_{4{}}^{2-}\)
\(\left[\mathrm{Ag}^{+}\right]=2 .24 \) ×10^-4mol L^-1
\(\mathrm{C_2O}_{4{}}^{2-} = \frac {2.24 \times 10 ^{-4}}{2}\) mol L^-1
= 1.12 ×10^-4mol L^-1
K_sp = [Ag]² [C₂O₄^2-]
= (2.24 ×10^-4mol L^-1) (1.12 ×10^-4mol L^-1)
= 5.619 × 10^-12mol³L^-3

Given that \(\alpha=1.20\)%=\(\frac{1.20}{100}\times1.2\times10^{-2}\)
\(K_{a}=\alpha^{2}c\)
\(=(1.2\times10^{-2})^{2}(0.1)=1.44\times10^{-4}\times10^{-1}\)
=\(1.44\times10^{-5}\)

\(\mathrm{Hg}_{2} \mathrm{Cl}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Hg}_{2}^{2+} \text { (aq) }+2 \mathrm{Cl^-}_{(\mathrm{aq})}\\ s \quad \quad \quad \quad \quad \quad s \quad \quad \quad \quad \quad 2s\)
\(\mathrm{K}_{\mathrm{sp}} =\left[\mathrm{Hg}_{2}^{2+}\right]{\left[\mathrm{Cl}^{-}\right]^{2}}  \)
\(=(\mathrm{s})(2 \mathrm{~s})^{2} \)
\(\mathrm{~K}_{\mathrm{sp}} =4 \mathrm{~s}^{3}\)

K_sp = 1.8 \(\times\)10^-10, [AgNO₃]= 1 M
\(\mathrm{AgCl}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \)
                    s            s
\(\mathrm{AgNO}_{3(\mathrm{aq})} \rightleftharpoons \mathrm{Ag}_{(\text {aq })}^{+}+\mathrm{NO}_{3_{(\text {aq })}}^{-}\\ 1 \mathrm{M} \quad \quad \quad \quad \quad 1 \mathrm{M} \quad \quad 1 \mathrm{M} \)
\(\left[\mathrm{Ag}^{+}\right]=(\mathrm{s}+1) \approx 1 \quad(\therefore \mathrm{s}<<1) \)
\(\left[\mathrm{Cl}^{-}\right]=\mathrm{s} \)
\(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \)
\(1.8 \times 10^{-10}=(1)(s) \)
\(\therefore \mathrm{s}=1.8 \times 10^{-10} \mathrm{M}\)

(i) p^H = -log₁₀ [H₃O⁺]
(ii) The pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.

[H₃O⁺] = 4 \(\times\) 10^-5M
pH = - log₁₀[H₃O⁺]
pH=-log₁₀[4 \(\times\) 10^-5]
pH = -log₁₀[4] - log₁₀[10^-5]        log₁₀ 10 = 1
pH = -log 4 + 5log₁₀10
= 5 - log 4
= 5 - 0.6021
=4.3979
Since pH is less than 7, the solution is acidic.

\(\mathrm{HClO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{ClO}_{4}^{-}\)
Bronsted Acid Proton Conjugate Base
HClO₄ can donate a proton. Therefore HClO₄ is an acid. Its conjugate base is ClO₄^-

When two are more solution are mixed, the resulting concentrations are different from the original.
\(\text { Molarity }=\frac{n}{\mathrm{~V}} \text { (or) } \mathrm{n}=\text { Molarity } \times \mathrm{v}  \)
Total Volume of the mixture = 0.15 + 0.1
= 0.25 L
\(\underset{0.1M}{Pb(NO_{3})_{2}}\rightleftharpoons \underset{0.1M}{Pb^{2+}}+2\underset{0.2M}{2NO^{-}_{3}}\)
n_Pb²⁺ \(=0.1\times0.15=0.015 \ mol\)
\([Pb^{2+}]_{mix}= \frac{n}{v} = \frac{0.1\times0.15}{0.25}=0.06M\)
\(\underset{0.2M}{NaCl}\rightleftharpoons \underset{0.2M}{Na^{+}}+\underset{0.2M}{Cl^{-}}\)
\(\mathrm{n}_{\mathrm{Cl^-}}=0.2 \times 0.1=0.02 \mathrm{~mol} \)
\(\left[\mathrm{Cl}^{-}\right]_{\text {mix }}=\frac{0.02}{0.25}=0.08 \mathrm{M} \)
\(\therefore Ionic \ Product =\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2} \)
\(=0.06 \times(0.08)^{2} \)
\(IP =3.84 \times 10^{-4}\)
\(\therefore 3.84 \times 10^{-4}>1.2 \times 10^{-5}\)
(or) \(\mathrm{IP}>\mathrm{K}_{\mathrm{sp}}\)
\(\therefore\) PbCl₂ will be precipitated.

\(h=\sqrt{K_{h}}=\sqrt{\frac{K_{w}}{K_{a}K_{b}}}=\sqrt{\frac{1\times10^{-14}}{1.8\times10^{-5}\times{1.8\times10^{-5}}}}\)
\(= \frac{1 \times10^{-7}}{1.8\times10^{-5}}\)
=\(0.7453\times10^{-2}\)
\(pH=\frac{1}{2}pK_{w}+\frac{1}{2}pK_{a}-\frac{1}{2}pK_{b}\)
Given that \(K_{a}=K_{b}=1.8\times10^{-5}\)
if K_a = K_b, then, pK_a = pK_b 
\(\therefore pH= \frac{1}{2}pK_{w}=\frac{1}{2}(14)=7\)
pH = 7

\(\text { Normality }=\text { Molarity } \times \text { Basicity }  \)
\(=0.04 \times 1  \)
\({\left[\mathrm{H}_{3} \mathrm{O}\right]^{+}=0.04=4 \times 10^{-2} } \)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]  \)
\(=-\log \left[4 \times 10^{-2}\right]  \)               log₁₀ 10 =1
\(=-\left[\log _{10} 4+\log _{10} 10^{-2}\right]  \)
\(=-\left[\log _{10} 4-2 \log _{10} 10\right]=2-\log _{10} 4  \)
= 2 - 0.6021 
= 1.3979 \(\simeq\) 1.40