Question

What volume of 0.623 M KOH is required to neutralize 23.5 mL if a 0.514 M H2SO4 solution? Bance the equation for the reaction.H2SO + KOH arrow K2SO4 + H2O

Answer 1

38.8 mL

Step-by-step 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:

`H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O`

Step 2: Calculate the volume of the base, KOH.

The volume of KOH can be determine using:

`(C_aV_a)/(n_a)=(C_bV_b)/(n_b)`

From the balanced equation, na = 1 and nb = 2. so putting the values of the given parameters into the above formula, we have:

`\begin{gathered} (0.514M*0.0235L)/(1)=(0.623M* V_b)/(2) \\ \\ V_b*0.3115M=0.012079M.L \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }0.3115M \\ \\ V_b=(0.012079M.L)/(0.3115M) \\ \\ V_b=0.0388\text{ }L \\ \\ The\text{ }volume\text{ }in\text{ }mL\text{ }is \\ V_b=0.0388*1000mL=38.8\text{ }mL \end{gathered}`

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