Question

4.0 gram of NaOH and 4.9 gram of H2SO4 are dissolved in water and volume is made upto 250ml. the ph of this solution is​

Answer 1

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Answer 2

Answer: The pH of the solution will be 7

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For NaOH:

Given mass of NaOH = 4.0 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{4.0g}{40g/mol}=0.1mol$

• For sulfuric acid:

Given mass of sulfuric acid = 4.9 g

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfuric acid}=\frac{4.9g}{98g/mol}=0.05mol$

Volume of the flask = 250 mL

The chemical equation for the reaction of NaOH with sulfuric acid follows:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

For neutralization:

2 moles of $H^+$ ion will react with 1 mole of $OH^-$ ion

Moles of $H^+$ ion = 2 × 0.05 = 0.1 moles

Moles of $OH^-$ ion = 0.1 moles

As, moles of $H^+$ ion = Moles of $OH^-$ ion

So, the pH of the solution will be 7