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Sagot :
Answer
38.8 mL
Explanation
Given:
Molarity of KOH, Cb = 0.623 M
Volume of H2SO4, Va = 23.5 mL = 0.0235 L
Molarity of H2SO4, Ca = 0.514 M
What to find:
The volume of KOH required to neutralize the acid.
Step-by-step Solution:
Step 1: Balance the equation for the reaction.
The balanced equation for the reaction is:
[tex]H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O[/tex]Step 2: Calculate the volume of the base, KOH.
The volume of KOH can be determine using:
[tex]\frac{C_aV_a}{n_a}=\frac{C_bV_b}{n_b}[/tex]From the balanced equation, na = 1 and nb = 2. so putting the values of the given parameters into the above formula, we have:
[tex]\begin{gathered} \frac{0.514M\times0.0235L}{1}=\frac{0.623M\times V_b}{2} \\ \\ V_b\times0.3115M=0.012079M.L \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }0.3115M \\ \\ V_b=\frac{0.012079M.L}{0.3115M} \\ \\ V_b=0.0388\text{ }L \\ \\ The\text{ }volume\text{ }in\text{ }mL\text{ }is \\ V_b=0.0388\times1000mL=38.8\text{ }mL \end{gathered}[/tex]Therefore, the volume of 0.623 M KOH is required to neutralize 23.5 mL if a 0.514 M H2SO4 solution = 38.8 mL
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