Tag: introduction to ionic equilibria in solution

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Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Generally speaking, how can you determine the charge of an ion formed by a representative element?

  1. The charge of the ion formed is related to the element's group number.

  2. The charge of the ion is related to the element's period number.

  3. An ion's charge is always 1/2 the atomic number.

  4. The charge of the ion is the number of valence electrons minus the number of core electrons.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Representative elements : s-block and p-block elements are representative elements.
The charge of the ion formed is related to the element's group number.
For example:
As Group-1 elements have +1 charge.
As Group-2 elements have +2 charge.

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

$\frac { N } { 10 }$ acetic acid was titrated with $\frac { N } { 10 }$ NaOH.When $25 \% , 50 \%$ and $75$$\%$ of titration is over then the pH of the solution will be $: \left[ \mathrm { K } _ { a } = 10 ^ { - 5 } \right]$

  1. $5 + \log 1 / 3,5,5 + \log 3$
  2. $5 + \log 3,4,5 + \log 1 / 3$
  3. $5 - \log 1 / 3,5,5 - \log 3$
  4. $5 - \log 1 / 3,4,5 + \log 1 / 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
t     CH_3COOH         NaOH          CH_3COO^-Na^+

0       0.1                      0.1
    
25%   0.1-0.025        0.1-0.025         0.025

50%   0.1-0.050        0.1-0.050         0.050

75%   0.1-0.075         0.1-0.075          0.075

$pH=- \log K _a+\log \dfrac{[salt]}{[acid]}$

t= 25%

$pH=5+\log \dfrac{0.025} {0.075}$

$pH=5+\log \dfrac{1} {3}$

t= 50%

$pH=5+\log \dfrac{0.050} {0.050}$

$pH=5+\log 1=5$

t= 75%

$pH=5+\log \dfrac{0.075} {0.025}$

$pH=5+\log 3$
Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Two electrolytic cells containing molten solutions of Nickel chloride and Aluminium chloride are connected in series. If same amount of electric current is passed through them, what will be the weight of Nickel obtained when $18gm$ of Aluminium is obtained? $\left( Al-27gm/mole,Ni-58.5gm/{ mole }^{ -1 } \right) $

  1. $58.5gm$
  2. $117gm$
  3. $29.25gm$
  4. $5.85gm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$By Faraday's Second Law$
$\frac{(m)N _{i}}{(m) _{N _{A1}}}=\frac{(E)N _{i}}{(E) _{N _{A1}}}$
$\frac{(m)N _{i}}{18}=\frac{58.5\times 2}{3\times 27}$
$(m){N _{i}}=58.5 g$

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

100 mL of 1 M HCl is mixed with 50 mL of 2 M HCl. Hence, $[H 3O^+]$ is _______.

  1. 1.00 M

  2. 1.50 M

  3. 1.33 M

  4. 3.00 M
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

FInal concentration of $H _3O^+,[H _3O^+]$=$\cfrac {V _1S _1+V _2S _2}{V _1+V _2}$

                                                                 =$\cfrac {100 \times 1+ 50 \times 2}{100 + 50}$
                                                                 =$ 1.33M$ .

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases
What concentrations of $CH _3COOH$ and $CH _3COONa$ are needed to prepare a 0.10M buffer at pH 5.0?

  1. 0.09

  2. 0.06

  3. 0.6

  4. 0.9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For acetic acid/acetate buffer at pH 5.0, using Henderson-Hasselbalch: pH = pKa + log([base]/[acid]). pKa of acetic acid = 4.76. So 5.0 = 4.76 + log([base]/[acid]), giving log([base]/[acid]) = 0.24, so [base]/[acid] = 1.74. With total [base] + [acid] = 0.10 M: let [acid] = x, then [base] = 1.74x, so x + 1.74x = 0.10, x = 0.0365 M (acid) and [base] = 0.0635 M. Option B (0.06) matches the base concentration.

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Solubility of $MX _{ 2 }$ type electrolytes is $0.5\times 10^{ -4 } mol/L$, Then find out ${ K } _{ sp }$ of electrolytes.

  1. $5\times 10^{ -12 }$
  2. $25\times 10^{ -10 }$
  3. $1\times 10^{ -13 }$
  4. $5\times 10^{ -13 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For an MX2 electrolyte, Ksp = [M^2+][X^-]^2 = (s)(2s)^2 = 4s^3. Given s = 0.5 * 10^-4, Ksp = 4 * (0.5 * 10^-4)^3 = 4 * 0.125 * 10^-12 = 0.5 * 10^-12 = 5 * 10^-13.

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Diamagnetism is exhibited by_____

  1. cobalt

  2. water

  3. oxygen

  4. iron
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Diamagnetic materials, like water, or water-based materials, have a relative magnetic permeability that is less than or equal to 1, and therefore a magnetic susceptibility less than or equal to 0. Diamagnetic materials are repelled by magnetic fields.

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

$H _2O \longrightarrow H^+ + OH^-$
The above reaction is dissociation or ionization?

  1. Dissociation

  2. Ionization

  3. Both dissociation and ionization together

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Ionization is the process by which an atom or a molecule acquires a negative or positive charge by gaining or losing electrons to form ions. It does not split like in dissociation reaction.

Therefore, $ { H } _{ 2 }O\rightarrow \quad { H }^{ + }+{ { OH }^{ - } } $ is an ionization reaction. 

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Which will not affect the degree of ionisation?

  1. Temperature

  2. Concentration

  3. Type of solvent

  4. Current
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Degree of ionisation changes according to the temperature, concentration and the type of electrolyte like strong electrolyte or weak electrolyte, it does not varies with current.