All categories

2 years ago

What does it accomplish

Chemistry

2 answers:

Hatshy [7]2 years ago

8 0

Answer:

wait what the question about

Explanation:

disa [49]2 years ago

4 0

What dose what accomplish??

You might be interested in

Consider the titration of a 20.0-mL sample of 0.105 M HC2H3O2 with 0.125 M NaOH. Determine each quantity. a. the initial pH b. t

Oksi-84 [34.3K]

Answer:

Explanation:

Given that:

Concentration of $HC_2H_3O_2 \ (M_1)$ = 0.105 M

Volume of $HC_2H_3O_2 \ (V_1)$ = 20.0 mL

Concentration of $NaOH (M_2)$ = 0.125 M

The chemical reaction can be expressed as:

$HC_2H_3O_2_{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O_{(l)}$

Using the ICE Table to determine the equilibrium concentrations.

$HC_2 H_3 O_2 _{(aq)} + H_2O _{(l) } \to C_2 H_3O_2^- _{(aq)} + H_3O^+_{ (aq)}$

I 0.105 0 0

C -x +x +x

E 0.105 - x x x

$K_a = \dfrac{[C_2H_5O^-_2][H_3O^+]}{[HC_2H_3O_2]}$

$K_a = \dfrac{(x)(x)}{(0.105-x)}$

Recall that the ka for $HC_2H_3O_2= 1.8 \times 10^{-5}$

Then;

$1.8 \times 10^{-5} = \dfrac{(x)(x)}{(0.105 -x)}$

$1.8 \times 10^{-5} = \dfrac{x^2}{(0.105 -x)}$

By solving the above mathematical expression;

x = 0.00137 M

$H_3O^+ = x = 0.00137 \ M \\ \\ pH = - log [H_3O^+] \\ \\ pH = - log ( 0.00137 )$

pH = 2.86

Hence, the initial pH = 2.86

b) To determine the volume of the added base needed to reach the equivalence point by using the formula:

$M_1 V_1 = M_2 V_2$

$V_2= \dfrac{M_1V_1}{M_2}$

$V_2= \dfrac{0.105 \ M \times 20.0 \ mL }{0.125 \ M}$

$V_2 = 16.8 mL$

Thus, the volume of the added base needed to reach the equivalence point = 16.8 mL

c) when pH of 5.0 mL of the base is added.

The Initial moles of $HC_2H_3O_2 =$ molarity × volume

$= 0.105 \ M \times 20.0 \times 10^{-3} \ L$

$= 2.1 \times 10^{-3}$

number of moles of 5.0 NaOH = molarity × volume

number of moles of 5.0 NaOH = $0.625 \times 10^{-3}$

After reacting with 5.0 mL NaOH, the number of moles is as follows:

$HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}$

Initial moles $2.1*10^{-3}$ $0.625 * 10^{-3}$ 0 0

F(moles) $(2.1*10^{-3} - 0.625 \times 10^{-3})$ 0 $0.625 \times 10^{-3}$ $0.625 \times 10^{-3}$

The pH of the solution is then calculated as follows:

$pH = pKa + log \dfrac{[base]} {[acid]}$

Recall that:

pKa for $HC_2H_3O_2=4.74$

Then; we replace the concentration with the number of moles since the volume of acid and base are equal

∴

$pH = 4.74 + log \dfrac{0.625 \times 10^{-3}}{1.475 \times 10^{-3}}$

pH = 4.37

Thus, the pH of the solution after the addition of 5.0 mL of NaOH = 4.37

We need to understand that the pH at 1/2 of the equivalence point is equal to the concentration of the base and the acid.

Therefore;

pH = pKa = 4.74

e) pH at the equivalence point.

Here, the pH of the solution is the result of the reaction in the $(C_2H_3O^-_2)$ with $H_2O$

The total volume(V) of the solution = V(acid) + V(of the base added to reach equivalence point)

The total volume(V) of the solution = 20.0 mL + 16.8 mL

The total volume(V) of the solution = 36.8 mL

Concentration of $(C_2H_3O^-_2)$ = moles/volume

= $\dfrac{2.1 \times 10^{-3} \ moles}{0.0368 \ L}$

= 0.0571 M

Now, using the ICE table to determine the concentration of $H_3O^+$ ;

$C_2H_5O^-_2 _{(aq)} + H_2O_{(l)} \to HC_2H_3O_2_{(aq)} + OH^-_{(aq)}$

I 0.0571 0 0

C -x +x +x

E 0.0571 - x x x

Recall that the Ka for $HC_2H_3O_2$ = $1.8 \times 10^{-5}$

$K_b = \dfrac{K_w}{K_a} = \dfrac{1.0\times 10^{-14}}{1.8 \times 10^{-5} } \\ \\ K_b = 5.6 \times 10^{-10}$

$k_b = \dfrac{[ HC_2H_3O_2] [OH^-]}{[C_2H_3O^-_2]}$

$5.6 \times 10^{-10} = \dfrac{x *x }{0.0571 -x}$

$x = [OH^-] = 5.6 \times 10^{-6} \ M$

$[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{5.6 \times 10^{-6} }$

$[H_3O^+] =1.77 \times 10^{-9}$

$pH =-log [H_3O^+] \\ \\ pH =-log (1.77 \times 10^{-9}) \\ \\ \mathbf{pH = 8.75 }$

Hence, the pH of the solution at equivalence point = 8.75

f) The pH after 5.09 mL base is added beyond (E) point.

$HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}$

Before 0.0021 0.002725 0

After 0 0.000625 0.0021

$[OH^-] = \dfrac{0.000625 \ moles}{(0.02 + 0.0218 ) \ L}$

$[OH^-] = \dfrac{0.000625 \ moles}{0.0418 \ L}$

$[OH^-] = 0.0149 \ M$

From above; we can determine the concentration of $H_3O^+$ by using the following method:

$[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{0.0149}$

$[H_3O^+] = 6.7 \times 10^{-13}$

$pH = - log [H_3O^+]$

$pH = -log (6.7 \times 10^{-13} )$

pH = 12.17

Finally, the pH of the solution after adding 5.0 mL of NaOH beyond (E) point = 12.17

3 0

3 years ago

When rocks return to the Earth's mantle, they can eventually melt and become magma. _______ and _______ are the main factors tha

gogolik [260]

When rocks returns to the Earth's mantle, they can eventually melt and become magma. Heat and Pressure are the main factors that controls this change

6 0

3 years ago

Read 2 more answers

What is the equilibrium expression for the reatcion bellow<br><br> 2SO3(g)<=>O2(g)+2SO2(g)

Luda [366]

Answer:

[O2(g)][SO2(g)]^2/[SO3(g)]^2

8 0

3 years ago

Which of the following equations does not demonstrate the law of conservation of mass?

enot [183]

The third option does not obey the law of conservation of mass.

Option 3.

Explanation:

The law of conservation of mass states that the sum of the masses of reactants should be equal to the sum of the masses of the products.

For example, if we consider the first option to verify if it obeys law of conservation of mass or not, 2 Na + Cl₂ → 2 NaCl

So one way to verify it is to find the mass of Na, then multiply it with 2, and then add this with 2 times of mass of chlorine. So this sum should be equal to the 2 times mass of NaCl. But it is somewhat lengthy.

Another way to easily determine this is to check if the elements are present equally in both sides. Such as, in reactant side and product side 2 atoms of Na is present . Similarly, the Cl atoms are also present in equal number in both reactant and product side. Thus this obeyed the law of conservation of mass.

Like this, if we see the second option, there also 1 atom of Na is present in reactant and product side and 2 molecules of H is present in reactant and product side, 1 oxygen is present in reactant and product side and 1 Cl is present in reactant and product side. So it also obeys the law of conservation of mass.

But in the third option, P₄ + 5 O₂→ 2 P₄O₁₀, here, there is 4 atoms of P in reactant side but in product side there is (4*2) = 8 atoms of P. Similarly, the number of atoms of oxygen in reactants and product side is also not same. So the third option does not obey the law of conservation of mass.

The fourth option also obeys the law of conservation of mass as the number of atoms of each element is same in both the product and reactant side.

Thus, the third option does not obey the law of conservation of mass.

5 0

3 years ago

What is used for mixing a small amount of chemicals together?

DENIUS [597]

a thin solid glass rod that is used in chemistry to combine substances. A stirring rod often has rounded ends and is about the length of a long straw.

<h3>What use serves the stirring rod?</h3>

A crucial component of lab apparatus for mixing chemicals and liquids for reactions is a long, thin stirring rod. Stirring rods are made of solid plastic, glass, or steel and are non-abrasive, chemically inert, and chemically resistant.

<h3>What is the name of the glass stirring rod?</h3>

Glass rod, also known as a stirring rod, stir rod, or solid glass rod, is frequently made of quartz and borosilicate glass. Its diameter and length can be modified to meet your needs.

<h3>Does filtration employ stirring rods?</h3>

When the liquid transfer procedure is paused, use a stirring rod to direct the liquid flow into the funnel and stop small amounts of liquid from dribbling down the beaker's outside.

learn more about stirring rod here

<u>brainly.com/question/9971891</u>

#SPJ4

7 0

2 years ago