Answer: The correct answer is Option d.
Explanation:
To calculate the molarity 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 KOH
We are given:
[tex]n_1=2\\M_1=0.543M\\V_1=72.1mL\\n_2=1\\M_2=?M\\V_2=39.0mL[/tex]
Putting values in above equation, we get:
[tex]2\times 0.543\times 72.1=1\times M_2\times 39.0\\\\M_2=2.01M[/tex]
Hence, the correct answer is Option d.
The concentration of the base is 2.01 M
A neutralization reaction is a reaction that occurs between an acid and a base to yield salt and water only;
This reaction occurs as follows; H2SO4 + 2KOH ----->K2SO4 + H2O
From the titration formula;
concentration of acid CA = 0.543 M
volume of acid VA = 72.1 mL
concentration of baseCB = ?
volume of base VB= 39.0 mL
Number of moles of acid = 1
Number of moles of base = 2
Given that;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CB = 0.543 M * 72.1 mL * 2/39.0 mL * 1
CB = 2.01 M
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