How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction is:H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l).

Question

asked Mar 25, 2020 139k views

1 Answer

Larry Smithmier · Answer 1 · 2020-03-30T14:32:23+0000

Answer : The volume of
required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), 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:

$n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?$

Putting values in above equation, we get:

$2* 0.300M* 85.0mL=1* 0.150M* V_2\\\\V_2=340mL$

Hence, the volume of
required to neutralize is, 340 mL