Tag: ionisation of weak acids and weak bases

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Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

$As _2S _3$ solution has negative charge, capacity to precipitate is highest in:

  1. $AlCl _3$
  2. $Na _3PO _4$
  3. $CaCl _2$
  4. $K _2SO _4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solution:- (A) $Al{Cl} _{3}$

According to Hardy-Schulze rule, more is the valence of effective ion, greater is its coagulating power.
Hence ${As} _{2}{S} _{3}$ precipitate the most in $Al{Cl} _{3}$.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

At $100^\circ C$, value of $K _{w}$ is 

  1. $1.0\times 10^{-14}\quad m^{2}$
  2. less than $1.0\times 10^{-14}\quad m^{2}$
  3. greater than $1.0\times 10^{-14}\quad m^{2}$

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

At higher temperature the value of $kw$ increases.This is in according with le-chatelier principle.

At $100^o kw=51.3\times 106{-14}$
C is the correct answer.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The solubility of CaF$ _2$ (K$ _{sp} = 5.3\times 10^{-9}$) in $0.1$ M solution of NaF would be : (Assume no reaction of cation/anion) .

  1. $5.3 \times 10^{-10}$ M
  2. $5.3 \times 10^{-8}$ M
  3. $5.3 \times 10^{-7}$ M
  4. $5.3 \times 10^{-11}$ M
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$(C)\ 5.3\times 10^{-7}m$

$CaF _2\rightleftharpoons Ca^{2+}+2F^-$

$K _{sp}=[Ca^{}2+][F^-]^2=S(S+0.1)^2=S\times 0.1^2=5.3\times 10^{-9}$

Note: $S<<0.1$ so, $S+ 0.1 \approx 0.1$ 

$\Rightarrow S=5.3\times 10^{-7}\ M$ 

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The percentage of pyridine $\left( {{C _5}{H _5}N} \right)$ that forms pyridinum ion $\left( {{C _5}{H _5}{N^ + }H} \right)$ in a $0.10M$ aqueous pyridine solution $\left( Given - {{K _b}, for \  {C _5}{H _5}N = 1.7 \times {{10}^{ - 9}}} \right)$ is    

  1. $0.0060\% $
  2. $0.013\% $
  3. $0.77\% $
  4. $1.6\% $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a weak base B, Kb = C * alpha^2 / (1 - alpha). Since alpha is very small, Kb = C * alpha^2. Here, 1.7e-9 = 0.1 * alpha^2, so alpha^2 = 1.7e-8, and alpha = 1.3e-4. The percentage is 1.3e-4 * 100 = 0.013%.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

A $40.0 ml$ solution of weak base, $BOH$ is titrated with $0.1 N - HCl$ solution. The $pH$ of the solution is found to be $10.0$ and $9.0$ after adding $5.0 ml$ and $20.0 ml$ of the acid, respectively. The dissociation constant of the base is ($log 2 = 0.3$)

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

Using the Henderson-Hasselbalch equation for a buffer (weak base + salt), pOH = pKb + log([salt]/[base]). After 5ml of HCl, 5ml of BOH is converted to B+, leaving 35ml BOH. After 20ml, 20ml BOH is converted to B+, leaving 20ml BOH. Solving the two equations for pKb yields 4.7, corresponding to Kb = 2e-5.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

If a salt of weak acid or base is added to a solution of its acid or base respectively, the:

  1. dissociation of acid or base is diminished
  2. the $pH$ of the solution in case of acid increases and in case of base decreases

  3. mixing of two leads for precipitation

  4. none of the above
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Due to common ion effect, if a salt of weak acid or base is added to a solution of its acid or base respectively, the dissociation of acid or base is diminished.


As a result, concentration of hydrogen ions or hydroxide ion will change and pH of solution increases in case of acid and decreases in case of base.

As $pH = -log [ H^+]$

Multiple choice chemistry chemical equilibrium and acids-bases dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

What will be the effect of adding $100 ml$ of $0.001 M - HCl$ solution to $100 ml$ of a solution having $0.1 M - HA$? The acid dissociation constant of $HA$ is $10^{-5}$.

  1. The degree of dissociation of $HA$ will decrease but the $pH$ of solution remains unchanged.
  2. The degree of dissociation of $HA$ remains unchanged but the $pH$ of solution decreases.
  3. Neither degree of dissociation nor $pH$ of solution will change.
  4. The degree of dissociation as well as $pH$ of solution will decrease.
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The common ion effects is shown by which of the following sets of solutions:-

  1. $BaCl _{2}+Ba(NO _{3}) _{2}$
  2. $NaCl+HCl$
  3. $NH _{4}OH+NH _{4}Cl$
  4. $None$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Common ion effect is normally applied in selective precipitation of ions which involves a technique of separating ions in an aqueous solution by using a reagent that precipitates one or more of the ions. addition of common ions in a solution that is already having that given ion normally leads to a formation of a precipitate. The common ion normally decreases the solubility of a slightly insoluble salt.

Its not (1) because they are both solids.

The remaining are solute solvent pairs. So all the other pairs will show common ion effect if the solute is soluble to some extent and exists in an equilibrium with the solvent. The answer given is (3).

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The value of observed and calculated molecular weight of silver nitrate are $92.64$ and 170 respectively. The degree of dissociation of silver nitrate is :

  1. $60\%$
  2. $83.5\%$
  3. $46.7\%$
  4. $60.23\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The van't Hoff factor i = (Calculated MW) / (Observed MW) = 170 / 92.64 = 1.835. For AgNO3, i = 1 + alpha(n-1). Since n=2, i = 1 + alpha. Thus, alpha = 0.835 or 83.5%.