Chemistry, 22.07.2019 03:00, jdkrisdaimcc11

A40.00 ml sample of 0.10 m weak acid with ka of 1.8×10−5 is titrated with a 0.10 m strong base. what is the ph after 20.00 ml of base has been added?

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answered: starlodgb1971

The answer is 4.74.
Solution:
The product of the volume and concentration of each solution gives the number of moles of the acid HA and the base OH- present before the neutralization:
0.0400L (0.10mol/L HA) = 0.00400 moles HA
0.0200L (0.10mol/L OH-) = 0.00200 moles OH-

We can see that the 0.00200 moles of the strong base OH- consumes 0.00200 moles of the weak acid HA:
HA + OH- → A- + H2O
initial 0.00400 0.00200 0
change -0.00200 -0.00200 +0.00200
equilibrium 0.00200 0 0.00200

The concentration of HA is equal to the concentration of A- after the reaction:
[HA] = [A-] = 0.00200mol / 0.0400L+0.0200L = 0.03333 M

The equilibrium-constant expression for HA is
Ka = [H+][A-] / [HA]
Ka = [H+] = 1.8x10^-5

We can now solve for pH:
pH = -log [H+] = -log(1.8x10^-5) = 4.74

answered: Guest

n2(g) + 3h2-> 2nh3(g) this is the balanced equation

note the mole ratio between n2, h2 and nh3. it is 1 : 3 : 2 this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note: the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

answered: Guest

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

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A40.00 ml sample of 0.10 m weak acid with ka of 1.8×10−5 is titrated with a 0.10 m strong base. what...