Tag: properties of acids and bases

Questions Related to properties of acids and bases

Find the $pH$ of the resulting solution and then mark the option in which $pH$ exists between color transition range of an indicator.

$50$ ml of $0.1$ M $HCO _3^- \ +$ $50$ ml of $0.8$ M $CO _3^{2-}$.

[For $H _2CO _3$ : $K _{a _1}=4\times 10^{-7}$ and $K _{a _2}=2\times 10^{-11}$]
  1. Phenol red (6.8 to 8.4)

  2. Propyl red 4(4.6 to 6.4)

  3. Phenolphthalein (8.3 to 10.1)

  4. Malachite green (11.4 to 13)


Correct Option: D
Explanation:
Given that
[Salt] = 0.8M   and  [Acid] = 0.1M

The solution can be regarded as acidic buffer solution containing weak acid sodium bicarbonate and its salt sodium carbonate with strong base.

The expression for the pH of the acidic buffer solution is as given below.

$pH=pK _a+log \frac {[salt]} {[acid]}$

$pK _a=-logK _a=-log [2 \times 10^{-11}]= 10.7$

Substitute values in the above solution.

$pH=10.7+log \frac {0.8} {0.1}=11.6$

Hence, the suitable indicator is Malachite green with pH range from 11.4 to 13.

When $20\ mL$ of $\dfrac {M}{10}NaOH$ are added to $10\ mL$ of $\dfrac {M}{10}HCl$, the resulting solution will:

  1. turn blue litmus red

  2. turn phenolphthalein solution pink

  3. turn methyl orange red

  4. have no effect on either red or blue litmus


Correct Option: B
Explanation:
$20$ ml of $\dfrac { M }{ 10 } NaOH=2\times { 10 }^{ -3 }$ moles of $NaOH$
$10$ ml of $\dfrac { M }{ 10 } HCl=1\times { 10 }^{ -3 }$ moles of $HCl$.
Excess amount of $NaOH=\left( 2\times { 10 }^{ -3 }-1\times { 10 }^{ -3 } \right) $ moles
                                             $=1\times { 10 }^{ -3 }$ moles
$\therefore$   $\left[ { OH }^{ - } \right] =\dfrac { { 10 }^{ -3 }\times 1000 }{ 30 } =0.033$
$\therefore$   $pH=14-pOH=12.49$
In this $pH$, Phenolphthalein solution turns into pink.
Answer will be $B$.

An indicator $HIn$ has a standard ionization constant of $9.0\times {10}^{-9}$. The acid colour of the indicator is yellow and the alkaline colour is red. The yellow colour is visible when the ratio of yellow form to red form is $30$ to $1$ and the red colour is predominant when the ratio of red form to yellow form is $2$ to $1$. What is the $pH$ range of the indicator?

  1. < $6.568$

  2. $6.568$ to $8.346$

  3. > $8.346$

  4. None of these


Correct Option: B
Explanation:
$ \underset {Yellow}{HIn} + H _2O \rightleftharpoons H _3O^+ + \underset {Red}{In^-}$
$K _{In}= (\dfrac {[H _3O^+][In^-]}{HIn})$
Yellow colour is visible when the ratio of acid form to base form is 3 to 1.

$K _{In}= (\dfrac {[H _3O^+][1]}{30})$
$9 \times 10^{-9}= (\frac {[H _3O^+][1]}{30})$
$[H _3O^+]= 270 \times 10^{-9}$
$-log [H _3O^+]= pH= 6.569$
Red colour is predominant when the ratio of base form to acid form is 2 to 1.

$K _{In}= (\dfrac {[H _3O^+][2]}{1})$
${[H _3O^+]}= 4.5 \times 10^{-9}$
$-log [H _3O^+]= pH= 8.523$
 The pH range of the indicator is 6.569 to 8.523.

Find the $pH$ of the resulting solution and then mark the option in which $pH$ exists between color transition range of an indicator.

50 ml of 0.2M HA solution $(K _a=10^{-5})$ + 50ml of 0.1M HCl solution + 100 ml of 0.13 M NaOH solution.

  1. Phenol red (6.8 to 8.4)

  2. Propyl red (4.6 to 6.4)

  3. Phenolphthalein (8.3 to 10.1)

  4. Malachite green (11.4 to 13)


Correct Option: B
Explanation:

Milimoles of HCl = Volume (ml) $\times$ Molarity = $50\times 0.1=5$ milimoles.
Milimoles of NaOH $=100 \times 0.13=13$ milimoles.

Some of the NaOH will be neutralized by HCl.
Milimoles of NaOH remaining unreacted $13-5=8$ milimoles.

Milimoles of HA$=50 \times 0.2 =10$ milimoles.
Out of this 8 milimoles will be neutralized to form a salt.

Thus the solution now contains 2 milimoles of weak acid HA and 8 milimoles of its salt with strong base.

It is an acidic buffer solution.
The expression for the pH of the acidic buffer solution is as given below.

$pH=pK _a+log \frac {[salt]} {[acid]}$

$pK _a=-log10^{-5}=5$

$pH=5+log \frac {8} {2}=5.6$.

Thus the suitable indicator is propyl red with pH range 4.6 to 6.4.

In the titration of $CH _{3}COOH$ against $NaOH$, we cannot use the

  1. Methyl orange

  2. Methyl red

  3. Phenolphthalein

  4. Bromothymol blue


Correct Option: A
Explanation:

For the titration of weak acid and strong bases like $CH _3COOH+NaOH$ methyl orange indicator is used to observe the end point.
Hence option A is correct.

During the titration of mixture of $NaOH,{ Na } _{ 2 }{ CO } _{ 3 }$ and an inert substance against hydrochloric acid:

  1. Phenolphthalein is used to detect the end point when $NaOH$ is completely neutralized and half of ${ Na } _{ 2 }{ CO } _{ 3 }$ is neutralized

  2. Methyl orange is used to detect the final end point

  3. Methyl orange is used to detect the first end point

  4. Phenolphthalein is used to detect the final end point


Correct Option: A,B
Explanation:

During the titration of mixture of $NaOH,{ Na } _{ 2 }{ CO } _{ 3 }$ and an inert substance against hydrochloric acid Phenolphthalein is used to detect the end point when $NaOH$ is completely neutralized and half of ${ Na } _{ 2 }{ CO } _{ 3 }$ is neutralized,but Methyl orange is used to detect the final end point.

Hence option A,B are correct.

(A) In general phenolphthalein is used as an indicator for the titration of weak acid $(CH _3COOH)$ and strong base (NaOH).
(R) At equivalence point solution is basic.

  1. Both (R) and (A) are true and reason is the. correct explanation of assertion

  2. Both (R) and (A) are true but reason is not correct explanation of assertion

  3. Assertion (A) is true but reason (R) is false

  4. Assertion (A) and reason (R) both are false

  5. Assertion (A) is false but reason (R) is true


Correct Option: A
Explanation:

If the titration is carried out for strong base and weak acid.At the end point the solution would be basic in nature .in this case phenolphthalein indicator.
So Phenolphthalein is used as an indicator for the titration of weak acid ($CH _3COOH$) and strong base ($NaOH$).Here the end point is basic in nature.

So Both (R) and (A) are true and reason is the. correct explanation of assertion.

Hence option A is correct.

The principal products obtained on heating iodine with concentrated caustic soda solution is:

  1. $NaIO\, +\, NaI$

  2. $NaIO\, +\, NaIO _3$

  3. $NaIO _3\, +\, NaI +\ H _2O$

  4. $NaIO _4\, +\, NaI$


Correct Option: C
Explanation:

On adding iodine to caustic soda(alkali), hypoiodites are formed and the peculiar smell is of iodoform and pale yellow in color.

$6NaOH + 3I _2 \longrightarrow NaIO _3 +5NaI +3H _2O$

Metallic oxides react with ............ to give salts and water, similar to the reaction of a/an ........... with an acid, metallic oxides are said to be basic oxides.

  1. bases, base

  2. acids, base

  3. acids, acid

  4. bases, acids


Correct Option: B
Explanation:

Metallic oxides react with acids to give salts and water, similar to the reaction of a base with an acid, metallic oxides are said to be basic oxides.

Choose the correct answer from the given option.
$Metal\ oxide + Acid \rightarrow $

  1. $Salt + Water$

  2. $Sweet + Water$

  3. $Light + Water$

  4. $Gas + Water$


Correct Option: A
Explanation:

The general reaction between a metal oxide and an acid can be written as : 

$Metal\ oxide + Acid \rightarrow Salt + Water$