Answer : The concentration of NaOH is, 0.336 M
Explanation:
To calculate the concentration 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=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL[/tex]
Putting values in above equation, we get:
[tex]2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL[/tex]
[tex]M_2=0.336M[/tex]
Thus, the concentration of NaOH is, 0.336 M
