Chemistry

Solutions and pH Chemistry

400 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 quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

How many grams of phosphoric acid $(H _2PO _4)$ would be needed to neutralise $100$g of magnesium hydroxide $(Mg(OH) _2)$?

  1. $66.7$ g
  2. $252$
  3. $112.6$ g
  4. $168$ g
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Apply the formula
${\left( {\dfrac{W}{Equivalent\ wt.}} \right) _{{H _3}P{O _4}}} = \left( {\dfrac{W}{Equivalent\ wt.}} \right) _{Mg{\left( {OH} \right) _2}}$
Hence,
${\dfrac{W}{{98 \times 3}} = \dfrac{{100}}{{58 \times 2}}}$
$\therefore{W = 112.6{\text{ }}gram}$
Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

How many grams of phosphoric acid would be needed to neutralize $100$ gm of magnesium hydroxide? (Molecular weight of $H _3PO _4=98$ and $Mg(OH) _2=58.3 gm$)

  1. 66.7 gm

  2. 252 gm

  3. 112 gm

  4. 168 gm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The balanced reaction is 2H3PO4 + 3Mg(OH)2 -> Mg3(PO4)2 + 6H2O. Calculate the moles of Mg(OH)2 (100/58.3), then use the stoichiometric ratio to find the required moles of H3PO4, and finally convert to grams.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

Which of the following is true :

  1. $pk _{b}$ for $OH^{-}$ is -1.74 at $25^{o}$C
  2. The equilibrium constant for the reaction between HA ($pk _{a} = 4$) and NaOH at $25^{o}$C will be equal to $10^{10}$
  3. The pH of a solution containing 0.1 M HCOOH ($k _{a} = 1.8 \times 10^{-4}$) and 0.1 M HOCN ($k _{a} = 3.2 \times 10^{-4}$) will be nearly (3 -log 7)

  4. All of the above are correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Acid strength and acid concentration represents:

  1. degree of dissociation and amount dissolved respectively

  2. amount dissolved and degree of dissociation respectively

  3. degree of dissociation and valency respectively

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

 Acid Strength refers to the degree of dissociation ( or ionization) in an aqueous solution. The greater the number of ions dissociated, or the number of cations and anions released in solution, the stronger the acid. Thus,  hydrochloric acid  dissociates completely into $H^+$ and $Cl^-$ ions in solution, so it is very strong. Acetic acid $(CH _3COOH)$,  dissociates feebly and releases few ions in solution, so it is  a weak acid.
Acid Concentration represents the amount of acid dissolved in a solvent. It is  measured in molarity ( the number of moles of acid in 1 L of acid solution), parts per million or percentage. The concentration is a ratio of the solute to solvent content of a solution. Acidic solutions with low numbers of acidic molecules/ions in solution are called dilute solutions whereas those with high numbers of acidic molecules/ions are called concentrated solutions.

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

The dissociation constants of monobasic acids A,B,C and D are $6 \times 10^{4}, 5 \times 10^{5}, 3.6 \times 10^{6}\ and\ 7 \times 10^{10}$ respectively. The pH values of their 0.1 molar aqueous solutions are in the order:

  1. A < B < C < D

  2. A > B > C > D

  3. A = B = C = D

  4. A > B < C > D
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As the dissociation constant increases (A<B<C<D), dissociation of acid increases (A<B<C<D) and hence $[H^+]$ increases  (A<B<C<D) and hence pH value  decreases  (A>B>C>D). 

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

The $K _a$ value for the acid $HA$ is $1.0 \times 10^{-6}$. What is the value of K for the  following reaction?
$A^+ + H _3O^+\rightleftharpoons HA + H _2O$

  1. $1.0 \times 10^{-8}$
  2. $1.0 \times 10^8$
  3. $1.0 \times 10^{-3}$
  4. $1.0 \times 10^6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$HA + H _2O\rightleftharpoons A^+ + H _3O^+$
$K _a=\dfrac{[A^-].[H _3O^+]}{[HA]}=10^{-6}$
As the given reaction is the reverse of above reaction $K$ will the reciprocal of the above reaction.
Therefore, $K=10^6$

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

Ionisation constant of each HA (weak acid) and BOH (weak base) are $3.0 x 10^{-7}$ each at 298K. The percentage degree of hydrolysis of BA at the dilution of 10L is :

  1. 25

  2. 50

  3. 75

  4. 40

  5. Data is insufficient
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\frac{h}{1-h}=\sqrt{KH}=\sqrt{\frac{kw}{K _{a}k _{b}}}=\sqrt{\frac{10^{-24}}{(3 \times 10^{-7})^{2}}}=\frac{1}{3}\Rightarrow h=0.25$

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases
When sulphuric acid dissolves in water, the following reactions take place:

$H _2SO _4\longrightarrow H^+ +HSO _4^-$  ($100\%$ ionisation)

$H _2SO _4^-\longrightarrow H^+ +SO _4^{2-}$     ($10\%$ ionisation)

If $0.2 M$ aqueous solution of $H _2SO _4$ was taken, the concentration of $[SO _4^{2-}]$ will be:
  1. $0.1$ M
  2. $0.01$ M
  3. $0.2$ M
  4. $0.02$ M
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$Step \: I: \: H _2SO _4\longrightarrow H^+ +HSO _4^-$;      $100$% ionisation

$Step \: II: \: HSO _4^-\longrightarrow H^+ +SO _4^{2-}$;     $10$% ionisation

     $H^++H _2O\longrightarrow H _3O^+$

$SO _4^{2-} \: is \: from \: step \: II$.

$Step \: I:  \: 100$% and hence, $[H^+]= 0.2 \: M$.

and  $[HSO _4^-]=0.2 \: M$.

$Step \: II :\:  \: 10$% and hence, $[H^+]=0.2 \times  0.1= 0.02 \:M$

and  $[SO _4^{2-}]= 0.02 \:M$.

Total $[H _3O^+]=0.2 + 0.02 =0.22 \:M$, 

$[SO _4^{2-}]= 0.02 \:M$ and $[HSO _4^-] =0.20-0.02 =0.18 \:M$.

Hence, the correct option is $(D)$
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

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.