29 - Chemical and Ionic Equilibrium Questions Answers

Amount of DCl = 4 milimole

Amount of NaOD = 8 milimole

After reaction of acid and base 

Amount of NaOD remaining = 4 milimole

Total volume = 100 ml = 0.1 lit

$$\begin{gathered} so \left[OD\right] =4*{10}^{-3}/0.1=4*{10}^{-2} \left(1\right) \\ Given pD = 13.6 \\ so \log \left[D^{+}\right] =-13.6= \overline{14}.4 \\ \Rightarrow \left[D^{+}\right] =2.5*{10}^{-14} \left(2\right) \\ Now solve \end{gathered}$$

question is not appropriate because for given K_w , [H⁺] will be 10^-3 and max. pH will be 6 and the question is asking about a solution of pH 7.5 which is out of limit 

but if it is considered as a correct question then answer will be (ii)

 N₂   +    O₂   =    2NO 

  1                       1                       0

  1-x                    1-x                  2x

so by formula K_c = (2x)² / (1-x)(1-x)

now solve for x

SO₂Cl₂ = SO₂ + Cl₂

_{100               0           0}

_{20                  40       40}

_{2 atm             4 atm   4 atm}

now calculate K_p 

N₂ (g) + 3H₂ (g) =  2NH₃ (g)

0.2          0.6                0

0.2-x       0.6-3x         2x

according to the given condition 4x = 0.8*40/100 = 0.32

so x = 0.08

now solve

A (g) + B (g) = C (g)

  1         1.5        2

1-x      1.5-x     2+x

according to the given condition 2+x = 0.5 so x = -1.5

now calculate K_c

Ag₂CrO₄ -------------> 2Ag + CrO₄

s                                            0                  0

                                              2s                 s

so K_sp = (2s)² s = 4s³

now solve   

Please submit complete question, K_a should be given with percentage of ionisation

For calculating pH of a salt, only concentration is not sufficient

This is a strong acid so consider 100 percent ionisation 

we get [H⁺] = 10²

it is more than 1 so pH = 0