# A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.235 M HCl and 0.605 M H2SO4. What volume of 0.19 M NaOH would be required to completely neutralize all of the acid in 273.1 mL of this solution?

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A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.235 M HCl and 0.605 M H2SO4. What volume of 0.19 M NaOH would be required to completely neutralize all of the acid in 273.1 mL of this solution?

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Step 1

Given:

Molarity of  HCl = 0.235 M

Molarity of H2SO4 = 0.605 M

Total volume = 273.1 mL

Molarity of NaOH = 0.19 M

We need to find what volume of 0.19 M NaOH would be required to completely neutralize all the acid.

Step 2

Chemical equation for HCl and NaOH reaction:

HCl + NaOH => NaCl + H2O

As per the above equation, equal moles of HCl and NaOH react with each other,

So,

Moles of NaOH = moles of HCl

Number of moles = volume of solution x concentration

Calculate the number of moles of HCl:

Number of moles of HCl  = volume of solution  x concentration of HCl

= 273.1 mL X 10 -3 L X 0.235 M

Moles of NaOH = moles of HCl = 64.17 X 10 -3 mol. Of HCl

Step 3

Similarly,

H2SO4 + 2 NaOH => Na2SO4 + 2 H2O

The ratio of H2SO4 and NaOH from the reaction is 1:2.

No. of moles of H2SO4 = volume of solution x concentration of H2SO4

= 273.1 mL X 10 -3 L X 0.605 M

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### Equilibrium Concepts 