Question

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)

Answer 1

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`