Chemistry, 31.03.2020 01:36, abbypoletick

When a 26.3 mL sample of a 0.465 M aqueous nitrous acid solution is titrated with a 0.461 M aqueous barium hydroxide solution, what is the pH after 19.9 mL of barium hydroxide have been added

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answered: HolliAnne

The pH of the solution after addition of barium hydroxide is 3.78

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .......(1)

For nitrous acid:

Molarity of nitrous acid = 0.465 M

Volume of solution = 26.3 mL

Putting values in equation 1, we get:

$0.465M=\frac{\text{Moles of nitrous acid}\times 1000}{26.3mL}\\\\\text{Moles of nitrous acid}=\frac{0.465\times 26.3}{1000}=0.0122mol$

For barium hydroxide:

Molarity of barium hydroxide = 0.461 M

Volume of solution = 19.9 mL

Putting values in equation 1, we get:

$0.461M=\frac{\text{Moles of barium hydroxide}\times 1000}{19.9mL}\\\\\text{Moles of barium hydroxide}=\frac{0.461\times 19.9}{1000}=0.0092mol$

The chemical reaction for nitrous acid and barium hydroxide follows the equation:

$2HNO_2+Ba(OH)_2\rightarrow Ba(NO_2)_2+2H_2O$

Initial: 0.0122 0.0092

Final: 0.003 - 0.0092

Volume of solution = 26.3+ 19.9 = 46.2 mL = 0.0462 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[NO_2^-]}{[HNO_2]}$

We are given:

pK_a = negative logarithm of acid dissociation constant of nitrous acid = 3.29

$[NO_2^-]=\frac{0.0092}{0.0462}$

$[HNO_2]=\frac{0.003}{0.0462}$

pH = ?

Putting values in above equation, we get:

$pH=3.29+\log(\frac{0.0092/0.0462}{0.003/0.0462})\\\\pH=3.78$

Hence, the pH of the solution after addition of barium hydroxide is 3.78

answered: Guest

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

answered: Guest

answer: atomic mass was originally used to arrange elements on the periodic table, but is no longer used to organize the modern version.

explanation:

originally atomic mass was used to arrange elements on the periodic table from left to right in a row. each column then shows similar properties of elements of a group.

in modern periodic table, the atoms are arranged in increasing atomic number. as atomic number represents the number of protons in an atom which is unique for every atom.