All categories

2 years ago

How to calculate the pH of NH4OH(0.1M)

1 answer:

8 0

Answer is: pH of ammonium hydroxide is 11.13.
Chemical reaction of ammonium hydroxide in water: NH₄OH → NH₄⁺ + OH⁻.

Kb(NH₄OH) = 1,8·10⁻⁵.
c₀(NH₄OH) = 0.1 M.
c(NH₄⁺) = c(OH⁻) = x.
c(NH₄OH) = 0.1 M- x.
Kb = c(NH₄⁺) · c(OH⁻) / c(NH₄OH).
0,000018 = x² / 0.1 mol/L - x.

Solve quadratic equation: x = c(OH⁻) = 0.00133 M.

pOH = -logc(OH⁻).

pOH = 2.87.

pH = 14 - 2.87.

pH = 11.13.

You might be interested in

This element has 80 electrons and is considered a transitional element. What am I?

sdas [7]

Answer:its mercury i think!!!!!

Explanation:

8 0

1 year ago

The density of osmium, which is the densest metal, is 22.57 g/cm^3. What is the mass of a block of osmium that measures 1.00 cm

olganol [36]

Density = Mass / Volume
V = 1.00 * 4.00 * 2.50 = 10 cm³
22.57 g/cm³ = Mass / 10 cm³
M = 22.57 g/cm³ * 10 cm³
M = 225.7 g
Answer: The mass of the block of osmium is 225.7 g.

7 0

2 years ago

Read 2 more answers

How many grams of fluorine must be reacted with excess lithium iodide to produce 10.0 grams of lithium fluoride?

rewona [7]

Answer:
7.32 g of F₂

Solution:
The equation is as follow,

2 LiI + F₂ → 2 LiF + I₂

According to equation,

51.88 g (2 mole) of LiF is produced from = 37.99 g (1 mole) F₂
So,
10 g of LiF will be produced by = X g of F₂

Solving for X,
X = (10 g × 37.99 g) ÷ 51.88 g

X = 7.32 g of F₂

8 0

2 years ago

Can someone help me

Molodets [167]

Answer:

The longer it takes to orbit the sun.

Explanation:

6 0

2 years ago

Read 2 more answers

1) What mass of Na2CO3 is required to make 50cc of its seminormal solution?

love history [14]

Answer:

$m=1.325gNa_2CO_3$

Explanation:

Hello,

In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:

$eq-g=0.5eq-g/L*0.050L=0.025eq-g$

Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:

$mol=0.025eq-gNa_2CO_3*\frac{1molNa_2CO_3}{2eq-gNa_2CO_3}\\ \\mol=0.0125molNa_2CO_3$

Finally, the mass is computed by using its molar mass (106 g/mol)

$m=0.0125molNa_2CO_3*\frac{106gNa_2CO_3}{1molNa_2CO_3} \\\\m=1.325gNa_2CO_3$

Regards.

7 0

2 years ago