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 substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Brine has a density of 1.2 g/cc. 40 cc of it is mixed with 30 cc of water. The density of the solution is:

  1. 2.11 g/cc

  2. 1.11 g/cc

  3. 12.2 g/cc

  4. 20.4 g/cc
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ \rho  } _{ mixture }=\dfrac { { \rho  } _{ 1 }{ V } _{ 1 }+{ \rho  } _{ 2 }{ V } _{ 2 } }{ { V } _{ 1 }+{ V } _{ 2 } } $


               $=\dfrac { 1.2\times 40+1\times 30 }{ 70 } =\dfrac { 78 }{ 70 } $


${ \rho  } _{ mixture }=1.11gm/cc$

Multiple choice bio-chemistry digestion digestion in small intestine digestive glands and their secretions alimentary canal

The pH of human small intestine is around $7.5$ and the pH of large intestine can be $5.5$. As substances travel from the small intestine to larger intestine, what would happen to the $H^+$ ions concentration?

  1. Increases $20$ fold
  2. Increases $2$ fold
  3. Increases $10$ fold
  4. Increases $100$ fold
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

pH scale is not a linear scale like a centimetre or inch scale (in which two adjacent values have the same difference). It is a logarithmic scale in which two adjacent values increase or decrease by a factor of 10. As pH 7.5 and 5.5 differ by 2, the difference in H$^{+}$ ion concentration between them would be 10$^{2}$ = 100.

So, the correct option is 'Increases 100 fold'.

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

One litre hard water contains $12 mg\, Mg^{2+}$ milli-equivalent of washing soda required to remove its hardness is: 

  1. $1$
  2. $12.16$
  3. $1\times 10^{-3}$
  4. $12.16\times 10^{-3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The molar mass of Mg2+ is 24 g/mol. 12 mg is 0.5 millimoles. Since Mg2+ has a valency of 2, this corresponds to 1 milliequivalent. Washing soda (Na2CO3) reacts in a 1:1 equivalent ratio to remove the hardness.

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

A volume of 100 L of hard water requires 5.6 g of lime for removing temporary hardness. The temporary hardness in ppm of CaCO3CaCO3 is:

  1. 56

  2. 100

  3. 200

  4. 112
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Temporary hardness is amount of $ CaCO _{3} $ in grams present in $ 10^{6}\,ml/grams $ of water


$ Ca(HCO _{3}) _{2}+CaO\rightarrow 2CaCO _{3}+H _{2}O $

$ 56\,g  $                      $ 2\times 100 = 200\,g $

$ 5.6\,g $                      $ 20\,g $ 

20 g $ CACO _{3} $ present in 100 L $ H _{2}O $

$ = 100\times 10^{3} = 10^{5}\,ml $

Then $ 10^{6} $ ml water contains 200g $ CaCO _{3} $

$ \therefore $ 200 ppm is temporary hardness 

Hence, the correct option is $\text{C}$.

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

One litre of a samle of hard water contains 1 mg of $CaCl _2$ and 1 mg of $MgCl _2$.Then the total hardness in terms of parts of $CaCO _3$ per $10^8$ parts of water by mass is:

  1. 1.954 ppm

  2. 1.260

  3. 0.946

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

$1$ mole of $CaCl _2\equiv1$ mole of $CaCO _3\equiv1$ mole of $MgCl _2$

$\therefore$ $100mg$ $CaCO _3$ is produced by $95mg$ $MgCl _2$
$\therefore$ $1mg$ of $MgCl _2$ gives $\cfrac {100}{95}mg$  $CaCO _3=$ $1.05mg$ $CaCO _3$

Similarly, $1mg$ of $CaCl _2$ gives $\cfrac {100}{111}mg$ $CaCO _3=$ $0.90mg$ $CaCO _3$

$\therefore$ Total $CaCO _3$ per litre of water= $1.05+0.90=1.95$ $mg$
Weight of $1000ml$ of water= $10^3g$=$10^6mg$

$\therefore$ Total hardness in terms of parts of $CaCO _3$ per $10^6$ parts of water by weight = $\cfrac {1.95}{10^6}\times 10^6$
=$1.95$ $ppm$

Answer: (A) $1.954$ $ppm$

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

100 ml of tap water containing $Ca(HCO _3) _2$ was titrated with N/50 $HCl$ with methyl orange as indicator. If 30ml of $HCl$ were required, calculate the temporary hardness as part of $CaCO _3$ per $10^6$ parts of water.

  1. 150 ppm

  2. 300 ppm

  3. 450 ppm

  4. 600 ppm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Milli equivalent of $Ca(HCO _{3}) _{2}$= miili equivalent of $CaCO _{3}$= mili equivalent of $HCl$

$N _{1}V _{1}= N _{2}V _{2}$
$1000 \times w$/($100$/$2$)= $30$/$50$
$w$= $0.03$ g
For $100$ml the amount of $CaCO _{3} = 0.03$ g
For $10^{6}$ = $0.03 \times 10^{6}$/$100=300$ ppm

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

One litre of a sample of hard water contains 55.5 mg of $CaCl _2$ and 4.75 mg of $MgCl _2$. The total harness in terms of ppm of $CaCO _3$ is :

  1. 9 ppm

  2. 10 ppm

  3. 20 ppm

  4. none of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$55.5$ mg of $CaCl _2$ in 1 litre of water corresponds to $55.5$ ppm.


$4.75$ mg of $MgCl _2$ in 1 litre of water corresponds to $4.75$ ppm.

Also $1 CaCl _2 = 1 CaCO _3$

$111$ ppm $= 100$ ppm

Hence, $55.5$ ppm $=\dfrac {100}{111} \times 55.5=50$ ppm.

Also $1 MgCl _2 = 1 CaCO _3$

$95$ ppm $= 100$ ppm

Hence, $4.75$ ppm $=\dfrac {95}{111} \times 4.75=5$ ppm.

Hence, the total hardness will be $5$ ppm.

Multiple choice chemistry matter around us measurement of properties effect of temperature and pressure on states of matter general introduction: importance and scope of chemistry

Which of the following can be expressed as grams per milliliter?

  1. Boiling point

  2. Rate of reaction

  3. Molecular mass

  4. Molarity

  5. Density
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Boiling point is expressed in kelvin, molecular mass in kilograms, molarity in moles per litre, rate of reaction in moles per litre per second and density in grams per litre or milliltre.

Multiple choice chemistry matter around us measurement of properties effect of temperature and pressure on states of matter general introduction: importance and scope of chemistry

What is the correct unit for the following solution?

$0.5mg$ of arsenic dissolved in $1kg$ of solution. (All the symbols given are in their standard form)

  1. $w/w$
  2. $m$
  3. $M$
  4. $v/v$
  5. $ppm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

0.5 mg of arsenic dissolved in 1 kg of solution. Both of the solute and solvent are given in terms of mass.

 Therefore, Option a is correct.

Multiple choice chemistry matter around us measurement of properties effect of temperature and pressure on states of matter general introduction: importance and scope of chemistry

A solution of hydrochloric acid is required for an experiment. The experimenter must first prepare the solution for use by diluting $10.0 mL$ of the hydrochloric acid to create $200 mL$ of the solution.
Which of the following would be the poorest choice of glassware for measuring the solute for the dilution?

  1. A graduated cylinder

  2. A volumetric flask

  3. A burrette

  4. A volumetric pipette
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\text{Option A is correct.}$
$\text{A graduated cylinder,measuring cylinder or mixing cylinder is a common piece of laboratory equipment }$$\text{used to measure the volume of a liquid}$
Multiple choice chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

Ammonia evolved from the treatment of $0.30$g of an organic compound for the estimation of nitrogen was passed in $100$ mL of $0.1$M sulphuric acid. The excess of acid required $20$ mL of $0.5$M sodium hydroxide solution for complete neutralization. The organic compound is:

  1. thiourea

  2. benzamide

  3. urea

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

Organic compound+ $H _2SO _4\longrightarrow$

Using Kjeldhal Process to calculate the percentage of nitrogen organic compound
Let unreacted $0.1M(=0.2N)H _2SO _4=Vml$
$\therefore, 20ml$ of $0.5M$ $NaOH=Vml$ of $0.2g$ $NH _2SO _4$
$\therefore 20 \times 0.5=V\times 0.2$
or, $V=50ml$
used $H _2SO _4=100-50=50ml$
$\therefore,$ % nitrogen=$14NV/w$
where $N$=normality of $H _2SO _4$
$V$=Volume of $H _2SO _4$ used
% nitrogen=$1.4\times 0.5\times 50\times 0.3=46.67$ %
Urea=$NH _2CONH _2=28\times 100/60$
$=46.67$ %
Thus, the organic compound is urea.