how would you write equation to show how buffers H2CO3 and NaHCO3 behave when (a) HCl is added and (b) NaOH is added.

1 answer:

8 0

H2CO3 <---> H+ + HCO3-
NaHCO3 <---> Na+ + HCO3-

When acid is added in the buffer, the excess H+ of that acid reacts with HCO3- to form H2CO3, and due to this NaHCO3 dissociates into HCO3- to attain the equilibrium. and hence there is no net effect of H+ due to pH remain almost constant. when a base is added to the buffer, the OH- ion of base react eith H+ ion present in buffer, then to attain equilibrium of H+ ion, the H2CO3 dissociates to produce H+ ion, but now there is the excess of HCO3- due to which Na+ ion react with them to attain equilibrium of HCO3-. hence there is again no net change in H+ ion due to which pH remain constant.....

You might be interested in

An atom of calcium contains 20 protons how many electrons does it have

harina [27]

20.

Atomic number is equivalent to its protons and electrons. :)

4 0

3 years ago

Describe how naturally acidic rainwater can affect a mountain of limestone.

Hitman42 [59]

The naturally acidic rainwater will erode the limestone mountin over time

6 0

3 years ago

What happens when iron is placed in a solution containing copper ions?

alexgriva [62]

When an iron is dipped in Copper Sulphate

Solution this reaction between them and

copper sulphate change into blue color to light

green color. This show that iron is more

reactive then copper, it can to replace copper

from CuSO4 , CuSO4 is of blue color and

FeSO4 is light green color.

Hope it helps

8 0

3 years ago

What compound is formed when 2,2-dimethyloxirane (2-methyl-1,2-epoxypropane) is treated with ethanol containing sulfuric acid

vladimir1956 [14]

Answer:

2-ethoxy-2-methylpropan-1-ol

Explanation:

On this reaction, we have an "epoxide" (2-methyl-1,2-epoxypropane). Additionally, we have acid medium (due to the sulfuric acid $H_2SO_4$ ). The acid medium will produce the hydronium ion ( $H^+$ ). This ion would be attacked by the oxygen of the epoxide. Then a carbocation would be produced, in this case, the most stable carbocation is the tertiary one. Then an ethanol molecule acts as a nucleophile and will attack the carbocation. Finally, a deprotonation step takes place to produce 2-ethoxy-2-methylpropan-1-ol.

See figure 1

I hope it helps!

3 0

3 years ago

Esters are generally pleasant smelling compounds True False

Rzqust [24]

Answer:

True

Explanation:

Esters are generally pleasantly smelling compounds. In fact, the fragrance industry uses esters to produce perfume, as well as uses esters as an ingredient to produce synthetic flavours and cosmetics, all of which have unique and pleasant smells.

8 0

3 years ago