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Naily [24]
1 year ago
13

In the titration of 50. 0 mL of 0. 400 M HCOOH with 0. 150 M LiOH, how many mL of LiOH are required to reach the equivalence poi

nt
Chemistry
1 answer:
mart [117]1 year ago
8 0

The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL

<h3>Balanced equation </h3>

HCOOH + LiOH —> HCOOLi + H₂O

From the balanced equation above,

The mole ratio of the acid, HCOOH (nA) = 1

The mole ratio of the base, LiOH (nB) = 1

<h3>How to determine the volume of LiOH </h3>
  • Molarity of acid, HCOOH (Ma) = 0.4 M
  • Volume of acid, HCOOH (Va) = 50 mL
  • Molarity of base, LiOH (Mb) = 0.15 M
  • Volume of base, LiOH (Vb) =?

MaVa / MbVb = nA / nB

(0.4 × 50) / (0.15 × Vb) = 1

20 / (0.15 × Vb) = 1

Cross multiply

0.15 × Vb = 20

Divide both side by 0.15

Vb = 20 / 0.15

Vb = 133.3 mL

Thus, the volume of the LiOH solution needed is 133.3 mL

Learn more about titration:

brainly.com/question/14356286

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A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 42 g of water (with an
kogti [31]

The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:

-Qm = Qw

Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)

where C is the specific heat capacities of the materials.

We calculate as follows:

-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)

-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)

Cm = 1.717 -----> OPTION C

8 0
3 years ago
Water enters a 4.00-m3 tank at a rate of 6.33 kg/s and is withdrawn at a rate of 3.25 kg/s. The tank is initially half full.
olga55 [171]

Answer:

At the start of the process, the volume not occupied by the water is 2 m3

Explanation:

At the start of the process you have a half full tank. It means that also a half is empty (not occupied by water).

Since the volume is 4 m3, 2 m3 are full (occupied by water) and 2 m3 (not occupied by water).

The volume in time will be

V(t)=V_0+(f_i-f_o)*t\\\\V(t) = 2 +(6.33/1000-3.25/1000)*t=2+0.00308*t \, \, [m3]

8 0
3 years ago
A reaction mixture, in an organic solvent, contains an organic acid, an organic base, neutral organic compounds, and base contai
zheka24 [161]

Answer: option b. Neutral organic compounds

Explanation:

3 0
3 years ago
If 26.64 mL of NaOH solution is required to react completely with 0.440 g KHP, what is the molarity of the NaOH solution
dybincka [34]

Answer:

0.08087M

Explanation:

3 0
2 years ago
The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.
madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

3 0
3 years ago
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