Titration help?

Suppose that 0.3000g sample consist of 50% by weight NaOH and 50% by weight Na2CO3. (a) How many milliliters of 0.1000M HCI would be required for titration to the phenolphthalein end point? (b) How many milliliters of HCl would have been required had methyl orange indicator been used instead of phenolphthalein?
answers (a,51.65)(b, 65.80)

1 Answer
Feb 2, 2018

See explanation

Explanation:

Since we have the total of 0.3000g which consist of 50% of each compounds.

Hence each compounds has = (0.3000g)/2 = 0.15g

Converting the grams to moles..

n = m/(Mm)

Where;

n = "No of moles"

m = "mass"

Mm = "Molar mass"

NaOH = (0.15 g)/(40 gmol^-1) = 0.00375 mol

Na_2CO_3 = (0.15 g)/(106 gmol^-1) = 0.001415 mol

Recall; c = n/v

Where;

n = "No of moles"

v = "Volume"

c = "Concentration"

c = 0.1000M

Divide each mol per HCL concentration of (0.1 mol)/l to get the volume in liters

v = n/c = 0.00375/0.1000 = 0.0375

v = n/c = 0.001415/0.1000 = 0. 1415

Therefore;

(A) Add volumes of HCL used:

=> 0.0375 + 0. 1415 = 0.05165 L = 51.65 ml

(B) Add volumes of HCL used:

=> 0.0375 + (0. 1415)2 = 0.0658 L = 65.8 ml