Question

What volume in milliliters of a 0.101 m naoh solution is required to reach the equivalence point in the complete titration of a 10.0 ml sample of 0.132 m h2so4?

Answer 1

mole ratio is 1:2, 1 mol H2SO4 reacts with 2 mols NaOH
mols of H2SO4=0.132*10/1000=0.00132 mols
mols of NaOH= 0.00132*2=0.00264
Vol of NaOH=0.00264*1000/0.101=26.14 mm

Answer 2

Answer: The volume of NaOH comes out to be 26.14 mL.

Step-by-step explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1,M_1\text{ and }V_1` are the n-factor, molarity and volume of acid which is
`H_2SO_4`

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is NaOH.

We are given:

`n_1=2\\M_1=0.132M\\V_1=10mL\\n_2=1\\M_2=0.101M\\V_2=?mL`

Putting values in above equation, we get:

`2* 0.132* 10=1* 0.101* V_2\\\\V_1=26.14mL`

Hence, the volume of NaOH comes out to be 26.14 mL.