Find how many milliliters of NaOH should be used to reach the half-equivalence point during the titration of 20.00 mL 0.888 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54×10–5) with 0.425 M NaOH solution. Enter 2 decimal places.

The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

Explanation:

The chemical equation for the dissociation of butanoic acid follows:

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

Neglecting the negative value because concentration cannot be negative

To calculate the volume of base, we use the equation given by neutralization reaction:

where,

are the n-factor, molarity and volume of acid which is butanoic acid

are the n-factor, molarity and volume of base which is NaOH.

We are given:

Putting values in above equation, we get:

Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

the answer is barium and copper bromide