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 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.

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

$100\ ml$ of an aqueous solution contains $6.0\times {10}^{21}$ solute molecules. The solution is diluted to $1$ lit. The number of solute molecules present in $10\ ml$ of the dilute solution is:

  1. $6.0\times {10}^{20}$
  2. $6.0\times {10}^{19}$
  3. $6.0\times {10}^{18}$
  4. $6.0\times {10}^{17}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
100 ml solution diluted to 1 liters (1000) ml contains $6.0 \times 10^{21}$ solute molecular

No of molecules present in 10 ml

$ = \dfrac{10 \times 6.0 \times 10 ^{21}}{1000} = 6 \times 10 ^{19} $

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

When $1$ mole of a substance is present in $1$ L of the solution, it is known as :

  1. normal solution

  2. molar solution

  3. molal solution

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

Molar concentration is a measure of the concentration of a solute in a solution and its unit is mol L$^{-1}$. Molarity is a method to express the concentration of a solution. It is defined as the number of moles of solute dissolved per liter of solution. 


Hence, when $1$ mole of a substance is present in $1$ L of the solution, it is known as a molar solution.

Option B is correct.

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

10g of sodium hydroxide dissolved in 1 L of water to make _____ solution.

  1. $0.25 M$
  2. $0.5 M$
  3. $1 M$
  4. $1.5 M$
  5. $4 M$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
molar mass of sodium hydroxide $=$$40g/mole$
$Molarity = \dfrac{Mass\, of\, solute}{(Molar\, mass\, of\, the\, solute)\times(Volume\, of \, solution\, in\, litres)}$

So $Molarity$ $=$$10/(40\times1)$$=$$0.25M$
Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

How much water, in liters, must be added to 0.5 L of 6 M HCl to make it 2 M?

  1. 0.33

  2. 0.5

  3. 1

  4. 1.5

  5. 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Molarity =$\dfrac{Mass \,of\, the\, solute}{(Molar \,mass\, of\, the\, solute)\times {(Vol. of\, soln.\, in\, liters)}}$

Mass of solute will remain same before and after mixing water.
so 
       $M _1V _1$$=$$M _2V _2$
or 
       $V _2$$=$$6\times0.5/2$$=$$1.5L$
this is final total volume so water added $=$$1.5-0.5$$=$$1L$ 
Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

What is the molar mass of a non-ionizing solid if 10 g of this solid, dissolved in 100 g of water, formed a solution that froze at $-1.21^o C$?

  1. 0.65 g

  2. 65 g

  3. 130 g

  4. 154 g

  5. 265 g
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the freezing point depression formula: deltaTf = Kf * m. With Kf = 1.86, deltaTf = 1.21, and molality m = (10/M) / 0.1, solving for M gives approximately 154 g/mol.

Multiple choice chemistry acids and alkalis neutralisations in everyday life acids and bases in daily life neutralisation reaction

A 26 ml of $ N-Na _{2}CO _{3} $ solution is neutralized by the solutions of acids A and B in different experiments. The volumes of the acids A and B required were $10 ml$ and $40 ml$, respectively. How many volumes of A and B are to be mixed in order to prepare 1 litre of normal acid solution? 

  1. $179.4, 820.6$
  2. $820.6, 179.4$
  3. $500, 500$
  4. $474.3, 525.7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ N _{1}V _{1} (Na _{2}CO _{3}) = N _{2}V _{2} (A)$
$ N _{1}V _{1} (Na _{2}CO _{3}) = N _{2}V _{2}(B)$
$ N _{1} = 1 , V _{1} = 26ml, V _{2} = 10ml, V _{3} = 40ml$
Normality of A
$ N _{2} = \dfrac{N _{1}V _{1}}{V _{2}} = \dfrac{1 \times 26}{10} = 2.6 N $
Normality of B
$ N _{3}= \dfrac{N _{1}V _{1}}{V _{3}} = \dfrac{1 \times 26}{40} = 0.65 N $
if we mix A & B then
$ N _{1}V _{1} + N _{2} V _{2} = N _{3} (V _{1}+V _{2})$
$ 2.6 V _{1}+0.65V _{2} = 1 \times 1000 $
$ 2.6V _{1}+ V _{2} + 0.65 V _{2} = 1000 ...(1)$
$ V _{1}+V _{2} =1000 ...(2)$
multiply $eq^{n}$ (2) by 0.65 
$ 0.65 V _{1}+0.65V _{2} = 650 ...(3)$
$ V _{2} = 820.6 ml$
$ V _{1} = 1000 - 820.6 = 179.4 ml $
option "a" correct.

Multiple choice chemistry acids and alkalis neutralisations in everyday life acids and bases in daily life neutralisation reaction

What volume of $ 0.18 N - KMnO _{4} $ solution would be needed for complete reaction with 25 ml of $ 0.21 N - KNO _{2} $ in acidic medium ?

  1. $57.29 ml$
  2. $11.67 ml$
  3. $29.17 ml$
  4. $22.92 ml$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ KMnO _{4}+KNO _{2}$
$ 0.18 N $ 25 ml
V = ? 0.21 N
we know $ N _{1}V _{1}= N _{2}V _{2}$
$ 0.18 \times V _{1} = 0.21 \times 25 $
$ V _{1} = \dfrac{0.21 \times 25}{0.18 }$
$ \boxed{V _{1} = 29.17 ml}$
Answer option C