We appreciate your visit to A 50 0 mL sample of a 1 00 M solution of a diprotic acid text H 2 text A text K a1 1 0. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
A 50. 0 ml sample of a 1.00 M solution of a diprotic acid is titrated with 2 M NaOH. The minimum volume of 2.00 m NaOH needed to reach a pH of 10 is 37.5 mL.
The molarity = 1 M
The volume = 50 mL
The moles of the diprotic acid = molarity × volume in L
The moles of the diprotic acid = 1 × 0.050
The moles of the diprotic acid = 0.050 mol
At the equivalence point, the moles of the diprotic acid is equals to the moles of the base = 0.050 mol.
At the 1st half equivalence:
pH = pka1 = 6
At the 2nd half equivalence :
pH = pka2 = 10
When the half reaction will occurs when we add 0.025moles of NaOH :
The total moles of NaOH = 0.075 moles
The volume of NaOH = 0.075 / 2
= 0.0375 L
= 37.5 mL
To learn more about diprotic acid here
https://brainly.com/question/14982663
#SPJ4
Thanks for taking the time to read A 50 0 mL sample of a 1 00 M solution of a diprotic acid text H 2 text A text K a1 1 0. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
