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

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

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:

$2\times 0.132\times 10=1\times 0.101\times V_2\\\\V_1=26.14mL$

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

Answer 2

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