Consider the following titration for these three questions: 1.00 L of 2.00 M HCl is titrated with 2.00 M NaOH.

a. How many moles of acid are equal to one equivalent in this titration?
b. How many moles of HCl are found in solution at the halfway point of the titration?
c. How many liters of base will be needed to reach the equivalence point of the titration?

a. 1 mole of acid is equal to one equivalent.

b. 1.00 moles of HCl are found.

c. 1L of 2.00M NaOH is needed to reach the equivalence point

Explanation:

HCl reacts with NaOH as follows:

HCl + NaOH → NaCl + H2O

Where 1 mole of HCl reacts with 1 mole of NaOH. The reaction is 1:1

a. As the reaction is 1:1, 1 mole of acid is equal to one equivalent

b. The initial moles of HCl are:

1.00L * (2.00moles HCl / 1L) = 2.00 moles of HCl

At the halfway point, the moles of HCl are the half, that is:

1.00 moles of HCl are found

c. At equivalence point, we need to add the moles of NaOH needed for a complete reaction of the moles of HCl. As the moles of HCl are 2.00 and the reaction is 1:1, we need to add 2.00 moles of NaOH, that is:

2.00moles NaOH * (1L / 2.00mol) =

1L of 2.00M NaOH is needed to reach the equivalence point

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.

where are the choice options?