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39 votes
39 votes
25.0 mL of 0.138M HCl were used to completely neutralize 39.0 mL of KOH. What is the concentration of the KOH used in this lab?

User Arkar Aung
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1 Answer

15 votes
15 votes

Answer:

0.0885M

Explanations:

In order to determine the concentration of the KOH used in this lab, we will use the dilution formula expressed as


C_1V_1=C_2V_2

where:

• C₁ and C₂ are the ,initial and final concentrations

,

• V₁ and V₂ are the ,initial and final volumes

Given the following parameters

• C₁ = 0.138M

,

• V₁ = 25.0mL = 0.025L

,

• V₂ = 39.0mL = 0.039L

Required

Final concentration C₂

Substitute the given parameter into the formula to have:


\begin{gathered} C_2=(C_1V_1)/(V_2) \\ C_2=(0.138*0.025)/(0.039) \\ C_2=(0.00345)/(0.039) \\ C_2=0.0885M \end{gathered}

Hence the concentration of the KOH used in this lab is 0.0885M

User Nick Orton
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