Chemistry

Solutions and pH Chemistry

451 Questions

Solutions and pH chemistry focus on the acidity or alkalinity of various substances, from human urine to acidic soils. Questions involve titration calculations, molarity determination, and understanding mole fractions. This topic is a staple in the chemistry sections of state and national level competitive exams.

Calculating solution molarityMole fraction calculationpH value interpretationAcidic soils and pHTitration and neutralization

Solutions and pH Chemistry Questions

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

One litre of water contains ${ 10 }^{ -7 }$ mole of ${H}^{+}$ ions. Degree of ionisation of water is:

  1. $1.8\times { 10 }^{ -7 }$
  2. $0.8\times { 10 }^{ -9 }$
  3. $5. 4\times { 10 }^{ -9 }$
  4. $5 . 4\times { 10 }^{ -7 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$H _2O⇌H^{ + }   +       OH^{ - }\quad $

  $C$        $0$                  $0$
$C(1-\alpha )$        $C\alpha $    $C\alpha $

$C\alpha=10^{-7}$

[H2O] =$\dfrac{ 1000}{18}$= 55.55 M 

C= 55.5M

So $\alpha=18\times10^{-10}$

In percentage $\alpha=1.8\times10^{-7}$%

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 weak acid, $HA$ has a ${K} _{a}$ of $1.00\times { 10 }^{ -5 }$. If $0.1$ mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closet to:

  1. $1$%
  2. $99.9$%
  3. $0.1$%
  4. $99$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
0.1 mole of acid is dissolved in 1 litre of water means $[HA]=0.1M$
Let '$\alpha$' be degree of dissocition
$HA\rightleftharpoons { H }^{ + }+{ A }^{ - }$
 $0.1$
$0.1(1-\alpha)$   $0.1\alpha$     $0.1\alpha$
${ K } _{ a }=\cfrac { \left[ { H }^{ + } \right] \left[ { A }^{ - } \right]  }{ \left[ HA \right]  } =\cfrac { { 0.1 }^{ 2 }{ \alpha  }^{ 2 } }{ 0.1(1-\alpha)  } $
Let $\alpha<<1$ so $1-\alpha=1$
$K _a=0.1\alpha^2=10^{-5}$
$\alpha=10^{-2}$
% of acid dissociated=$1$%
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 concentration of two acids are some, their relative strengths can be compared by:

  1. ${ \alpha } _{ 1 }/{ \alpha } _{ 2 }$
  2. $K _{ 1 }/K _{ 2 }$
  3. ${ \left[ { H }^{ + } \right] } _{ 1 }/{ \left[ { H }^{ + } \right] } _{ 2 }$
  4. $\sqrt { K _{ 1 }/K _{ 2 } } $
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

Relative strength of two acids can be compared by their degree of dissociation.

$HA\rightleftharpoons H^++A^-$
$C$
$C-C\alpha$   $C\alpha$    $C\alpha$
If concentration of two acids are same so their relative strength can be compared by their $[H^{+}]$ concentration.
$K _a=C\alpha^2$
$\alpha=(K _a/C)^{0.5}$
If concentration of two acids are same so their relative strength can be compared by square root of their dissociation constants.

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

Determine $[{OH}^{-}]$ of a $0.050\ M$ solution of ammonia to which has been added sufficient ${NH} _{4}Cl$ to make the total $[{NH} _{4}^{+}]$ equal to $0.100 M$. $[{K} _{b({NH} _{3})}=1.8\times {10}^{-5}]$

  1. $[{OH}^{-}]=9.0\times {10}^{-6}$
  2. $[{OH}^{-}]=9.0\times {10}^{-8}$
  3. $[{OH}^{-}]=9.0\times {10}^{-2}$
  4. $[{OH}^{-}]=9.0\times {10}^{-9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${NH} _{4}Cl\longrightarrow {NH} _{4}^{+}+{Cl}^{-}$
${NH} _{4}OH\longrightarrow {NH} _{4}^{+}+{OH}^{-}$
${K} _{b}=\cfrac { \left[ { NH } _{ 4 }^{ + } \right] \left[ OH \right]  }{ \left[ { NH } _{ 4 }OH \right]  } $
$[{NH} _{4}^{+}]=$ is due to salt because ${NH} _{4}OH$ ionise less amount due to common ions effect
$1.8\times {10}^{-5}=\cfrac{0.1\times [{OH}^{-}]}{0.05}$ 
$9\times {10}^{-6}=[{OH}^{-}]$

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 $pH$ of a dilute solution of acetic acid was found to be $4.3$ The addition of a small crystal of sodium acetate will cause $pH$ to:

  1. become less than $4.3$
  2. become more than $4.3$
  3. remain equal to $4.3$

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

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

Due to this common ion effect, when we add sodium acetate dissociation of acetic acid decreases and solution will have less number of hydrogen ion and so, pH increases. (as $pH = -log [H^+]$)

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 monoprotic acid in $1.00M$ solution is $0.001$% ionised. The dissociation constant of acid is:

  1. ${ \alpha }^{ 2 }C+\alpha K-K=0$
  2. ${ \alpha }^{ 2 }C-\alpha K-K=0$
  3. ${ \alpha }^{ 2 }C-\alpha K+K=0$
  4. ${ \alpha }^{ 2 }C+\alpha K+K=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to Ostwald's dilution law


$K=\cfrac { { \alpha  }^{ 2 }C }{ \left( 1-\alpha  \right)  } $

On solving, we get


$\alpha^2C+\alpha K-K=0$


Hence, the correct option is $\text{A}$

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 ionisation constant of acetic acid is $1.8\times { 10 }^{ -5 }$.The concentration at which it will be dissociated to $2$% is:

  1. $1M$
  2. $0.045M$
  3. $0.018M$
  4. $0.45M$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When acetic acid is dissolved in water, it partially dissociates (2%).

Ionisation constant $ K _a=1.87\times 10^{-5}$
$K _a=\cfrac{C\alpha \times C\alpha}{C(1-\alpha)}$

We assume $\alpha\sim  0\Rightarrow 1-\alpha\sim 1$
$\Rightarrow 1.87\times 10^{-5}=\cfrac{C\alpha^2}{1}={C\times 0.02\times 0.02}$
$\Rightarrow C = \cfrac{1.87\times 10^{-5}}{0.02^2}$
$=0.0467M$
Correct answer is option-B.

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 dissociation constants of two acids $ H{ A } _{ 1 }$ and $H{ A } _{ 2 }$ are $3.0\times { 10 }^{ -4 }$ and $1.8\times { 10 }^{ -5 }$ respectively. The relative strengths of the acids will be:

  1. $1:4$
  2. $4:1$
  3. $1:16$
  4. $16:1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The dissociation constants of $HA _1$ and $HA _2$ are $3\times10^{-4} $ and $1.8\times10^{-5}.$

The strength of an acid is directly proportional to square root of dissociation constants of acids. So relative strength of the given acids are:
$\dfrac { { (Acidic\ Strength }) _{ HA _1 } }{ { (Acidic\ Strength }) _{ HA _2} } =\dfrac { \sqrt { { (Dissociation\ Constant }) _{ HA _1} }  }{ \sqrt { { (Dissociation\ Constant }) _{ HA _2 } }  } $

 Relative Acidic Strength= $\dfrac { { (Acidic Strength }) _{ HA _1 } }{ { (Acidic Strength }) _{ HA _2 } } =\dfrac { \sqrt { 3.0\times { 10 }^{ -4 } }  }{ \sqrt { 1.8\times { 10 }^{ -5 } }  } =4.08=4(approx)$

relative strengths of acids will be $4:1.$
Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

What is the $[OH^-]$ in the final solution prepared by mixing $20.0\ mL$ of $0.050\ M$ $HCl$ with $30.0\; mL$ of $0.10 \;M\; Ba(OH) _2$?

  1. $0.12\ M$
  2. $0.10\ M$
  3. $0.40\ M$
  4. $0.0050\ M$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$Ba(OH) _2 + 2HCl \rightarrow BaCl _2 + 2H _2O$

2 m mol of HCl neutralize 1 m mole of $Ba(OH) _2$

1 m mol of HCl neutralize 0.5 m mol of $Ba(OH) _2$

$Ba(OH) _2$ left = 3 - 0.5 m mol = 2.5 m mol

         $[Ba(OH) _2] = \frac{2.5}{50}\;M = 0.05\; M$

or      $[OH]^- = 2 \times 0.05  = 0.1\; M$
Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

 8 gram of NaOH is mixed with 9.8 gram of $H _{2}SO _{4}$, the pH of the solution is:

  1. more than 7

  2. 7

  3. less than 7

  4. cant be said
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

8 gram (0.2 mole) of NaOH (molecular weight 40 g/mol) completely neutralizes 9.8 gram (0.1 mole) of $H _2SO _4$ (molecular weight 98 g/mol).
Since the molar concentration of both the compound are approximately same, the resulting solution will be neutral. Its pH will be 7.

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

The ionisation constant of an acid, $K _a$ is the measure of strength of an acid. The $K _a$ values of acetic acid, hypochlorous acid and formic acid are $1.74 \times 10^{-5}, 3.0 \times 10^{-8}$ and $1.8 \times 10^{-4}$ respectively. Which of the following orders of pH of $0.1$mol $dm^{-3}$ solutions of these acids is correct?

  1. Acetic acid > Hypochlorous acid > Formic acid

  2. Hypochlorous acid > Acetic acid > Formic acid

  3. Formic acid > Hypochlorous acid > Acetic acid

  4. Formic acid > Acetic acid > Hypochlorous acid
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$K _a$ is a measure of the strength of the acid i.e., larger the value of $K _a$, the stronger is the acid.
Thus, the correct order of acidic strength is
$HCOOH > CH _3COOH > HClO$
Stronger the acid, lesser will be the value of pH. Hence, the correct order of pH is $HClO > CH _3COOH > HCOOH$.

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

The $[H^+]$ of a resulting solution that is $0.01\space M$ acetic acid $(K _a = 1.8\times10^{-5})$ and $0.01\space M$ in benzoic acid $(K _a = 6.3\times10^{-5})$:

  1. $9\times10^{-4}$
  2. $81\times10^{-4}$
  3. $9\times10^{-5}$
  4. $2.8\times10^{-3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a mixture of two weak acids, [H+] = sqrt(Ka1*C1 + Ka2*C2). [H+] = sqrt(1.8e-5 * 0.01 + 6.3e-5 * 0.01) = sqrt(0.81e-6) = 0.9e-3 = 9e-4 M.