Which element has the largest atomic radius

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

[tex]HN_3 + OH- ---> N^-_{3} + H_2O[/tex]

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of [tex]HN_3[/tex] = [tex]\dfrac{0.001755}{0.0383}[/tex] = 0.0458 M

Concentration of [tex]N^{-}_3[/tex] = [tex]\dfrac{ 0.001995 }{0.0383}[/tex] = 0.0521 M

GIven that :

Ka = [tex]1.9 x 10^{-5}[/tex]

Thus; it's pKa = 4.72

[tex]pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})[/tex]

[tex]pH =4.72 + log(1.1376)[/tex]

[tex]pH =4.72 + 0.05598[/tex]

[tex]pH =4.77598[/tex]

pH ≅ 4.80

The pH of the solution 0.150 M hydrazoic acid after 13.3 mL of NaOH base is added is 4.80.

pH of the given solution will be used by using the following equation:

pH = pKa + log[conjugate base] / [weak acid]

Given chemical reaction will be represented as:

HN₃ + OH⁻ → N₃⁻ + H₂O

Moles will be calculated as:

n = M×V, where

M = molarity

V = volume

Moles of 0.150 M hydrazoic acid = (0.150M)(0.025L) = 0.00375  mol

Moles of 0.150 M NaOH = (0.0133)(0.150) = 0.001995 mol

From the above calculation it is clear that moles of hydrazoic acid is present in excess and it will be:

0.00375 - 0.001995 = 0.001755 mol

And 0.001995 mol of N₃⁻ is preduced by the reaction.

Total volume of the solution = 0.025 +  0.0133 = 0.0383  L

To calculate the pH after titration, first we have to calculate the concentration in terms of molarity of N₃⁻ and HN₃ as:

[N₃⁻] = 0.001995 mol / 0.0383  L = 0.0521 M

[HN₃] = 0.001755 mol / 0.0383  L = 0.0458 M

Ka for HN₃ = 1.9 × 10⁻⁵

pKa = -log( 1.9 × 10⁻⁵ ) = 4.72

On putting all these values on the above equation, we get

pH = 4.72 + log (0.0521) / (0.0458)

pH = 4.80

Hence, pH of the solution is 4.80.

To know more about pH, visit the below link:
https://brainly.com/question/10313314

Answer:

A.Electrolysis

Explanation:

A.Electrolysis

For example,  electrolysis of solution of NaCl in water gives H2 and O2.

Answer:

please look at the picture below.

Explanation: