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4 years ago

What is the PH of a 0.05M solution of the strong base Ca(OH)2 at 25C

Chemistry

1 answer:

krok68 [10]4 years ago

3 0

Answer:

pH = 14 - pH = 14 - 1 = 13

Explanation:

0.05M Ca(OH)₂ => 0.05M Ca⁺(aq) + 2(0.05M) OH⁻(aq)

pOH = -log[OH⁻] = -log(0.100) = 1

pH = 14 - pH = 14 - 1 = 13

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What mass of magnesium would combine with exactly 16.0 grams of oxygen

TEA [102]

1.65g MgO = 1g Mg
1.65 - 1 = 0.65 g of O in MgO

solve it using proportion:
1g Mg / 0.65g O = x (g) Mg / 16g O
or 1 / 0.65 = x / 16

24.6 g is the answer.

if 1 gram of oxygen requires 1.65 grams of Mg
then 16 grams of oxygen will require 16 ( 1.65) or 26.4 grams.

3 0

4 years ago

What is the mole fraction of NaOH in an aqueous solution that

astraxan [27]

Answer:

The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

Explanation:

We are given that

Aqueous solution that contains 22.9% NaOH by mass means

22.9 g NaOH in 100 g solution.

Mass of NaOH(WB)=22.9 g

Mass of water =100-22.9=77.1

Na=23

O=16

H=1.01

Molar mass of NaOH(MB)=23+16+1.01=40.01

Number of moles = $\frac{Given\;mass}{Molar\;mass}$

Using the formula

Number of moles of NaOH $(n_B)=\frac{W_B}{M_B}=\frac{22.9}{40.01}$

$n_B=0.572moles$

Molar mass of water=16+2(1.01)=18.02g

Number of moles of water $(n_A)=\frac{77.1}{18.02}$

$n_A=4.279 moles$

Now, mole fraction of NaOH

= $\frac{n_B}{n_B+n_A}$

$=\frac{4.279}{0.572+4.279}$

=0.882

Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

4 0

3 years ago

I would like some assistance

charle [14.2K]

Answer:

#2.

Explanation:

Look at the charges. Both are positive, therefore both are cations.

6 0

3 years ago

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Verdich [7]

Third choice is correct.

6 0

3 years ago

If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M TRIS base, what will be the resulting pH?

Katyanochek1 [597]

Answer: The pH of resulting solution is 8.7

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For TRIS acid:

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol$

For TRIS base:

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

$0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol$

Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of TRIS acid = 8.3

$[\text{TRIS acid}]=\frac{0.005}{0.11}$

$[\text{TRIS base}]=\frac{0.012}{0.11}$

pH = ?

Putting values in above equation, we get:

$pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7$

Hence, the pH of resulting solution is 8.7

6 0

4 years ago

What is the PH of a 0.05M solution of the strong base Ca(OH)2 at 25C​

What is the PH of a 0.05M solution of the strong base Ca(OH)2 at 25C