Answer: The molarity of KOH is 0.84 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 KOH.
We are given
[tex]n_1=2\\M_1=1.50M\\V_1=25.2mL\\n_2=1\\M_2=?M\\V_2=90mL[/tex]
Putting values in above equation, we get:
[tex]1\times 1.50\times 25.2=2\times M_2\times 90\\\\M_2=0.84M[/tex]
Hence, the molarity of KOH is 0.84 M.
