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Blizzard [7]
1 year ago
6

24. 00 ml of a 0. 25 m naoh solution is titrated with 0. 10m hcl. What is the ph of the solution after 24. 00 ml of the hcl has

been added?.
Chemistry
1 answer:
12345 [234]1 year ago
7 0

pH of the solution after 24. 00 ml of the hcl has been added is 12.87

millimoles NaOH = mL x M = 24.00 mL x 0.25 M = 6.00

millimoles HCl = 24.00 mL x 0.10 M = 2.40

total volume = 48.00 mL

.................................NaOH + HCl ==>NaCl + H2O

initial.........................6.00.........0............0.........0

added.....................................2.40............................

change.................... -2.40......-2.40.........+2.40.... +2.40

equilibrium.................3.60.........0..............2.40.......2.40

The NaCl contributes nothing to the pH of the final solution. The pH is determined by the excess of NaOH present. (NaOH) = millimoles/mL = 3.60/48.00 = 0.075 M = (OH^-)

pOH = -log (OH^-). Then

pOH = -log (0.075)

pOH =1.1249

As we know,

pH + pOH = pKw = 14.00

pH=14-pOH

pH=14-1.1249

pH=12.87

<h3>What is pH?</h3>

pH is a logarithmic measure of an aqueous solution's hydrogen ion concentration. pH = -log[H+], where log is the base 10 logarithm and [H+] is the concentration of hydrogen ions in moles per liter.

The pH of an aqueous solution describes how acidic or basic it is, with a pH less than 7 being acidic and a pH greater than 7 being basic. A pH of 7 is regarded as neutral (e.g., pure water). pH values typically range from 0 to 14, though very strong acids may have a negative pH and very strong bases may have a pH greater than 14.

Learn more about pH:

brainly.com/question/491373

#SPJ4

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Explanation:

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How many grams N2F4 can be produced from 225 g F,?​
zavuch27 [327]

Answer:

308 g

Explanation:

Data given:

mass of Fluorine (F₂) = 225 g

amount of N₂F₄ = ?

Solution:

First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄

Reaction:

          2F₂ + N₂ -----------> N₂F₄

Now look at the reaction for mole ratio

          2F₂     +    N₂   ----------->  N₂F₄

        2 mole                              1 mole

So it is 2:1 mole ratio of Fluorine to N₂F₄

As we Know

molar mass of F₂ = 2(19) = 38 g/mol

molar mass of N₂F₄ = 2(14) + 4(19) =

molar mass of N₂F₄ = 28 + 76 =104 g/mol

Now convert moles to gram

                 2F₂          +       N₂   ----------->  N₂F₄

        2 mole (38 g/mol)                        1 mole (104 g/mol)

                 76 g                                           104 g

So,

we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.

Apply unity formula

                  76 g of F₂ ≅ 104 g of N₂F₄

                   225 g of F₂ ≅ X of N₂F₄

Do cross multiplication

                  X of N₂F₄ = 104 g x 225 g / 76 g

                  X of N₂F₄ = 308 g

So,

308 g N₂F₄ can be produced from 225 g F₂

7 0
3 years ago
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