Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

A 0.200 M NaOH solution was used to titrate a 18.25 mL HE
solution. The endpoint was reached after 31.20 mL of titrant
were added. Find the molar concentration of the original HF
solution.

Sagot :

Neetoo

Answer:

M₂ = 0.34 M

Explanation:

Given data:

Molarity of NaOH =M₁ =  0.200 M

Volume of HF =V₂=  18.25 mL

Volume of NaOH added = V₁ = 31.20 mL

Molarity of HF solution = M₂ = ?

Solution:

M₁V₁  =  M₂V₂

by putting values,

0.200 M × 31.20 mL = M₂ × 18.25 mL

M₂ = 0.200 M × 31.20 mL /  18.25 mL

M₂ = 6.24 M.mL / 18.25 mL

M₂ = 0.34 M

Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.