Respuesta :
Answer:
0.823 M was the molarity of the KOH solution.
Explanation:
[tex]H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O[/tex] (Neutralization reaction)
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=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL[/tex]
Putting values in above equation, we get:
[tex]2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL[/tex]
[tex]M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M[/tex]
0.823 M was the molarity of the KOH solution.
