Chemistry, 30.03.2020 23:33 erikamaldonado661

Calculate the pH of a solution that contains 0.250 M formic acid, HCOOH (Ka =1.8 x 10-4 ), and 0.100M sodium formate, HCOONa after the addition of 10.0 mL of 6.00M NaOH to the original buffered solution volume of 500.0 mL.

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Answer from: nave1008

Answer : The pH of the solution is 3.5

Explanation :

First we have to calculate the moles of HCOOH , HCOONa and NaOH .

$\text{Moles of }HCOOH=\text{Concentration of }HCOOH\times \text{Volume of solution in L}$

$\text{Moles of }HCOOH=0.250M\times 0.500L=0.125mol$

and,

$\text{Moles of }HCOONa=\text{Concentration of }HCOONa\times \text{Volume of solution in L}$

$\text{Moles of }HCOONa=0.100M\times 0.500L=0.05mol$

and,

$\text{Moles of }NaOH=\text{Concentration of }NaOH\times \text{Volume of solution in L}$

$\text{Moles of }NaOH=6.00M\times 0.010L=0.06mol$

The equilibrium chemical reaction is:

$HCOOH+OH^-\rightleftharpoons HCOO^-+H_2O$

Initial moles 0.125 0.06 0.05

At eqm. (0.125-0.06) 0 (0.05+0.06)

= 0.185 = 0.11

Given:

$K_a=1.8\times 10^{-4}$

Now we have to calculate the value of pK_a .

The expression used for the calculation of pK_a is,

$pK_a=-\log (K_a)$

Now put the value of K_a in this expression, we get:

$pK_a=-\log (1.8\times 10^{-4})$

$pK_a=4-\log (1.8)$

pK_a=3.7

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[HCOONa]}{[HCOOH]}$

$pH=pK_a+\log \frac{[\frac{\text{Moles of HCOONa}}{\text{Volume of solution}}]}{[\frac{\text{Moles of HCOOH}}{\text{Volume of solution}}]}$

As the volume of solution are same. So,

$pH=pK_a+\log \frac{\text{Moles of HCOONa}}{\text{Moles of HCOOH}}$

Now put all the given values in this expression, we get:

$pH=3.7+\log (\frac{0.11}{0.185})$

pH=3.5

Therefore, the pH of the solution is 3.5

Answer from: Quest

Sugar and water are chemically combined.(a solution is chemically combined while a mixture is mixed together without bonding the molecules)

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

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Calculate the pH of a solution that contains 0.250 M formic acid, HCOOH (Ka =1.8 x 10-4 ), and 0.100...