Question

Calculate the pH of 0.01 M sulphuric acid ?​

Answer 1

★ To Find :-

• pH of 0.01 M sulphuric acid

★ As We Know That :-

$\large{\pink{\sf{☞ \: Formula \: of \: Sulphuric \: Acid = H _{2}SO_{4}}}}$

$\large{\pink{\sf{☞ \: One \: Mole \: of \: H _{2}SO_{4} \: gives \: 2 \: mole \: of \: {H}^{ + } \: ions.}}}$

$\large{\pink{\sf{☞ \: pH = \: -log[ {H}^{ + }] \: where \: {H}^{ + }\: refers \: to \: the \: Concentration \: of \: {H}^{ + }\: ions \: or \: the \: Hydronium \: ions.}}}$

Then,

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$\huge{\red{\boxed{\boxed{\green{\underline{\underline{\mathfrak{ ★ \: Answer :- }}}}}}}}$

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$\large{\blue{\sf{ [ {H}^{ + } ] \: ion \: concentration \: in \: 0.01M \: H _{2}SO_{4}}}}$

= 0.01 × 2

=0.02 mole/l

$\large\green{\underline{\red{\sf{ \therefore \: pH=−log[0.02]}}}}$

$\large\green{\underline{\red{\sf{ pH =1.699 }}}}$

which is your required answer!!

Answer 2

$\dag\:\underline{\sf Question:-}$

Calculate the pH of 0.01 M sulphuric acid.

$\dag\:\underline{\sf Answer:-}$

Given:-

• Concentration of sulphuric acid = 0.01 M

To find:-

• pH in 0.01 M of sulphuric acid

Formula to be used:-

• pH = - log₁₀[H₃O⁺]

Solution:-

[H₃O⁺] = 2 × Concentration

           = 2 × 0.01

           = 2 × 10⁻²

                                                                                 

Using the formula:-

pH = - log₁₀[H₃O⁺]

     = - log₁₀ (2 10⁻²)

     = - log₁₀2 - log₁₀10⁻²

     = - log₁₀2 + 2

     = 2 - 0.3010

     = 1.699

∴ The pH of 0.01 M sulphuric acid is 1.699

Hope this helps! ♡