Answer:- 0.138 M
Solution:- The buffer pH is calculated using Handerson equation:
acts as a weak acid and 
as a base which is pretty conjugate base of the weak acid we have.
The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:
Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.
So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.
Let's plug in the values in the equation:
Taking antilog:
On cross multiply:
[base] = 1.38(0.10)
[base] = 0.138
So, the concentration of the base that is required to make the buffer is 0.138M.
The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.
<span>So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Answer:
Moles of KOH in 1000 mL solution = 0.255 moles
Moles of KOH in 1 mL solution = 0.255/1000 = 0.000255 moles
Moles in 95 mL solution = (95 * 255)/1000000 = 24225/1000000
Moles of KOH in 95 mL 0.255M solution = 0.024225 moles
Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Solution : Given,
Volume of solution = 500 ml
Molarity of KOH solution = 0.189 M
Molar mass of KOH = 56 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of solute KOH.
Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
