Question

It takes 38 mL of 0.75 M NaOH solution to completely neutralize 155 mL of a sulfuricacid solution (H2SO4). a.Write a balanced equation for the neutralizationof NaOH with H2SO4

Answer 1

a.
`H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O`

b. 0.092 M

Step-by-step explanation:

Hello!

In this case, since for this titration process we know the used volume of 0.75-M sodium hydroxide needed to neutralize 155 mL of the acid, we first need to write the undergoing chemical reaction between them:

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

This, since there is 1:2 mole ratio between the acid and the base, at the equivalent point we must respect:

`2*n_(acid)=n_(base)`

That in terms of concentrations and volumes is:

`2*M_(acid)V_(acid)=M_(base)V_(base)`

Thus, the concentration of acid was:

`M_(acid)=(M_(base)V_(base))/(2*V_(acid)) \\\\M_(acid)=(0.75M*38mL)/(2*155mL) \\\\M_(acid)=0.092M`

Best regards!