Question

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Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Volume of NaOH = 10 ml.
• Molarity of NaOH = 0.45M.
• Volume of HCl = 40 ml.
• Molarity of HCl = 0.10M.

$\huge{\bold{\underline{Explanation:-}}}$

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As we know,

Volume of NaOH is 0.10 ml ,

converting it to Litres,

We get the volume as,

$\large{\underline{ \tt Volume \: of \: NaOH = 0.01 \: L}}$

Molarity = 0.45M.

Now,

As we know,

$\large{\tt Molarity = \dfrac{Moles}{Volume\: of \: solution_{(liters)}}}$

Substituting the values,

$\large{ \tt 0.45 = \dfrac{n}{0.01}}$

$\large{ \tt n = 0.01 \times 0.45}$

$\large{\boxed{ \tt n = 0.0045 \: moles}}$

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Volume of HCl is 0.40 ml ,

converting it to Litres,

We get the volume as,

$\large{\underline{ \tt Volume \: of \: HCl = 0.04 \: L}}$

Molarity = 0.1M.

Now,

As we know,

$\large{\tt Molarity = \dfrac{Moles}{Volume\: of \: solution_{(liters)}}}$

Substituting the values,

$\large{ \tt 0.1 = \dfrac{n}{0.04}}$

$\large{ \tt n = 0.004 \times 0.1}$

$\large{\boxed{ \tt n = 0.004 \: moles}}$

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Now, writing the chemical reaction,

$\large{\bold{ \tt NaOH + HCl \xrightarrow{} NaCl + H_2O}}$

Here we see ,

1 mole of NaOH reacts with HCl to neutralize the effect.

Now,

$\large{ \tt 1 \: mole \: of HCl = 1 mole \: of NaOH}$

$\large{ \tt 0.004 \: mole \: of HCl = 0.004 \: mole \: of NaOH}$

Therefore,

Sodium hydroxide will be left unreacted, and will cause the Ph effect,

As all other Sodium hydroxide has neutralized Hydrochloric acid.

No. of moles left (NaOH) = 0.0045 - 0.004

No. of moles left (NaOH) = 0.0005.

$\large{\boxed{ \tt No. \: of \: moles \: left \: (NaOH) = 0.0005}}$

As here there is domination of Basic nature then we have to find first p(oH).

$\large{ \tt concentration \: of \: OH^{-} \: ions = \dfrac{moles \: left}{total \: volume}}$

Substituting the values,

$\large{ \tt concentration \: of \: OH^- \: ions = \dfrac{0.0005}{0.01 + 0.04}}$

$\large{ \tt concentration \: of \: OH^- \: ions = \dfrac{0.0005}{0.05}}$

$\large{\boxed{ \tt concentration \: of \: OH^- \: ions = 0.01}}$

As we know the formula,

$\large{\boxed{\tt p(OH) = - log(OH)}}$

Substituting the values,

$\large{ \tt p(OH) = - log(10^{-2})}$

It comes as,

$\large{ \tt p(OH) = 2}$

But we know,

$\large{\boxed{ \tt pH + p(OH) = 14}}$

substituting ,

$\large{ \tt pH + 2 = 14}$

$\large{\tt pH = 14 - 2}$

$\huge{\boxed{\boxed{ \tt pH = 12 }}}$

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So, the pH of the solution is 12.

Therefore, option - 2 is correct.