Equilibrium

$\left[N{H}_{4}Cl\right]=\frac{2}{60}=\frac{1}{30}M$

$pH=7-\frac{1}{2}P{K}_b-\frac{1}{2}logC$

$=7-\frac{5}{2}-\frac{1}{2}log\left(\frac{1}{30}\right)=5.24$

Ka for C₃H₇COOH = 2 * 10^-5

$pKa=-\mathcal{l}og\left(2*10^{-5}\right)=5-\mathcal{l}og2$  

=5 – 0.3 = 4.7

pH of 0.2 (M) solution = 

$pH=\frac{pKa-\mathcal{l}ogC}{2}$  

$=\frac{1}{2}\left(4.7\right)-\frac{1}{2}\mathcal{l}og\left(0.2\right)$  

$pH=27*10^{-1}$     

Ans 27

$P_B^O=$ vapour pressure of benzene at 20°C = 70 torr

$P_M^O$ = vapour pressure of toluene at 20°C = 20 torr

Mixture is equimolar, X_B = 0.5 and X_M = 0.5

Total vapour pressure (P_T) = 70 * 0.5 + 20 * 0.5 = 45 torr

Mole fraction of benzene in vapour phase $\left(x_B^{\text{'}}\right)=\frac{P_B^O{X}_B}{P_T}=\left(\frac{70*0.5}{45}\right)torr$

= 0.777 = 77.7 * 10^-2 torr

Ans. = 78 (the nearest integer)

$\left[C{l}^{-}\right]=10^{-1}M$

$\left[Cr{O_4}^{--}\right]=10^{-3}M$

for AgCl ppt¹, ${\left[A{g}^{+}\right]}_{req}=\frac{1.7*10^{-10}}{10^{-1}}M$  

For Ag₂CrO₄ppt², ${\left[A{g}^{+}\right]}_{req}=\sqrt[]{\frac{1.9*10^{-12}}{10^{-3}}}M$

${\left[A{g}^{+}\right]}_{req}=4.3*10^{-5}M$

Being lower concentration of [Ag⁺] in case of AgCl, it will precipitate first.

50 ml of 1 (M) HCl + 30 ml of 1 (M) NaOH

NaOH    +            HCl  ->          NaCl + H₂O

30 * 1 mmol      50 * 1 mmol     

0 mmol               20 mmol

${\left[H^{+}\right]}_{mix}=\frac{20}{50+30}M=\frac{20}{80}M=\frac{1}{4}M=0.25M$

$x*10^{-4}=6021*10^{-4}$    

$\therefore$ x = 6021

Ans. = 6021

0.0504 M NH₄Cl of 5ml => millimole of $N{H}_4^{+}=0.0504*5$

0.0210 M NH₃ of 2ml => millimole of NH₃ = 0.0210 * 2

It is a basic buffer.

Total volume = 7ml

$Here,K_b=\frac{\left[O{H}^{-}\right]*\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}$     

$\therefore 1.8*10^{-5}=\frac{\left[O{H}^{-}\right]*0.0504*5}{0.0210*2}$         

$\therefore \left[O{H}^{-}\right]=\frac{1.8*10^{-5}*0.0210*2}{0.0504*5}=0.3*10^{-5}M$  

$\therefore \left[O{H}^{-}\right]=3*10^{-6}M$

$\therefore x=3$   

Ans. = 3

Statement I is true because methyl orange indicator has pH range (3.2 to 4.2) which is suitable for buffer produced by strong acid and weak base.

Statement II is false because phenolphthalein has pH range (8.4 to 10.2) and only suitable for equivalence point above than 7, In this case pH increased by the addition of NaOH on acetic acid.

Acid = M (OH)₂ ® Salt + H₂O

M.E of Acid = me of Base

$\therefore 10*\left(0.1*nf\right)=\left(0.05*2\right)*30$

$\therefore n_f=3$

Thus basicity of acid = 3

Total concentration of H+ after the mixing of HCl and H2SO4;

$\left[H^{+}\right]=\frac{\left(200*0.01\right)+\left(400*2*0.01\right)}{600}$

$\left[H^{+}\right]=\frac{1}{60}$

$\therefore pH=lo{g}_{10}\left(\frac{1}{60}\right)=1.78$

For an acid- base titration, Methly orange exist at end point as quinonoid form.