Question

for titrating,NaOH vs HCL, phenolphthalein and methyl orange, both are suitable indicators. Why?​

Answer 1

Answer:

may be it will help you

this is was the explanation

Answer 2

Answer:

Suppose 10 drops of solution from a burette make 1 cm³. Hence, the volume of 1/2 drop of solution is 0.05 cm².

Consider the titration of 25.00 cm³ of 0.100 M HCl (in the conical flask) against 0.100 M NaOH (from the burette).

When the titrant is 1/2 drop before the equivalence point, i.e. 24.95 cm³ of NaOH is added:

[H⁺] in the solution = (0.100 × 25 - 0.100 × 24.95)/(25.00 + 24.95) = 1.00 × 10⁻⁴ M

pH = -log[H⁺] = -log(1.00 × 10⁻⁴) = 4

When the titrant is 1/2 drop after the equivalence point, i.e. 25.05 cm³ of NaOH is added:

[OH⁻] in the solution = (0.100 × 25.05 - 0.100 × 25.00)/(25.05 + 25.00) = 1.00 × 10⁻⁴ M

pH = pKw + log[OH⁻] = 14 + log(1.00 × 10⁻⁴) = 10

Explanation:

Refer to the titration curve below. At the equivalence point the addition of one drop of base to the acid causes an increase of 6 pH units. The vertical portion of curve lies between pH 4 and pH 10. Both the pH ranges of phenolphthalein and methyl orange lie on this vertical portion and thus both of them are proper indicators.