H2SO4 + 2 NaOH ----> Na2SO4 + 2 H2O.
0.085 L * 0.176 mol/L = 0.01496 mol H2SO4
is neutralised by 0.01496 mol * 2
= 0.02992 mol NaOH.
1000 mL of 0.492 M NaOH
contains 0.492 moles NaPH.
0.02992 / 0.452 * 1000 mL
= 66.19 = 66 mL
Answer : The volume of [tex]NaOH[/tex] needed is, 45.4 mL
Explanation :
To calculate the volume of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=2\\M_1=0.176M\\V_1=85.0mL\\n_2=1\\M_2=0.452M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]2\times 0.176M\times 58.0mL=1\times 0.450M\times V_2\\\\V_2=45.4m:[/tex]
Hence, the volume of [tex]NaOH[/tex] needed is, 45.4 mL
