Ask question

Ask question

All categories

3 years ago

10

Solid calcium hydroxide is dissolved in water until the pH of the solution is 10.44. Determine the concentration of the calcium

hydroxide in the solution.

1 answer:

Orlov [11]3 years ago

6 0

Answer:

1.38x10⁻⁴ M

Explanation:

First, we <u>calculate the pOH of the solution</u>, using the <em>given pH</em>:

pOH = 14 - pH

pOH = 14 - 10.44

pOH = 3.56

Now we <u>calculate the concentration of OH⁻</u>, using<em> the pOH</em>:

pOH = -log[OH⁻]

[OH⁻] = $10^{-pOH}$ = 2.75x10⁻⁴ M

Following the reaction Ca(OH)₂ → Ca⁺² + 2OH⁻, we divide [OH⁻] by 2 to calculate [Ca(OH)₂]:

[Ca(OH)₂] = 2.75x10⁻⁴ M / 2

[Ca(OH)₂] = 1.38x10⁻⁴ M

You might be interested in

A sample was prepared by mixing 18. ml of 3.00 x 10^-3 m crystal violet (cv) with 2.00 ml of 0.250 m naoh. calculate the resulti

Aleks [24]

Answer : The resulting concentrations of CV and NaOH are 0.0027 M and 0.025 M respectively.

Explanation :

Step 1 : Find moles of crystal violet and NaOH.

The molarity formula is

$Molarity = \frac{mol}{L}$

Molarity of crystal violet = $3.00 \times 10^{-3} = \frac{mol (CrystalViolet)}{L}$

The volume of crystal violet solution is 18 mL which is 0.018 L.

Moles of crystal violet = $3.00 \times 10^{-3} \times 0.018 = 5.4 \times 10^{-5}$

Moles of crystal violet = 5.4 x 10⁻⁵

Moles of NaOH = $Molarity \times L = 0.250 \times 0.00200 = 5.00 \times 10^{-4}$

Moles of NaOH = 5.00 x 10⁻⁴

Step 2 : Find total volume of the solution

The total volume of the solution after mixing NaOH and crystal violet is

0.018 L + 0.00200 = 0.020 L

Step 3 : Use molarity formula to find final concentrations

Molarity of crystal violet = $\frac{mol(CrystalViolet)}{Total Volume(L) } = \frac{5.4 \times 10^{-5}}{0.020} = 2.7 \times 10^{-3}$

Final concentration of CV = 0.0027 M

Molarity of NaOH= $\frac{mol(NaOH)}{Total Volume(L) } = \frac{5.00 \times 10^{-4}}{0.020} = 0.025 \times 10^{-3}$

NaOH is a strong base and dissociates completely as follows.

$NaOH (aq) \rightarrow Na^{+} (aq) + OH^{-} (aq)$

The mole ratio of NaOH and OH⁻ is 1:1 . Therefore the concentration of OH⁻ is same as that of NaOH.

Concentration of OH⁻ = 0.025 M

8 0

3 years ago

For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow

lora16 [44]

Answer:

Rate constant of the reaction is $3.3\times 10^{-3} M^{-2} s^{-1}$ .

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

$R=k[A]^a[B]^b[C]^c$

Rate(R) of the reaction in trail 1 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.30 M$

$R=9.0\times 10^{-5} M/s$

$9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c$ ...[1]

Rate(R) of the reaction in trail 2 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.90 M$

$R=2.7\times 10^{-4} M/s$

$2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c$ ...[2]

Rate(R) of the reaction in trail 3 ,when :

$[A]=0.60 M,[B]=0.30 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c$ ...[3]

Rate(R) of the reaction in trail 4 ,when :

$[A]=0.60 M,[B]=0.60 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c$ ...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

$R=k[A]^2[B]^0[C]^1$

Rate constant of the reaction = k

$9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1$

$k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}$

$k=3.3\times 10^{-3} M^{-2} s^{-1}$

7 0

3 years ago

Determine the volume of fluid in the graduated cylinder shown.

alexgriva [62]

Answer:

47.3 ml

Explanation:

The graduated cylinder is shown in the image attached.

Now we have to take a good look at the cylinder, the lines between 45 and 50 are 46, 47, 48 and 49. Even though the points in between two lines weren't graduated but we can intelligently guess the correct volume by observing the upper meniscus of the liquid. Hence the answer.

8 0

3 years ago

Read 2 more answers

This is for anyone who’s good with science coz I absolutely hate it. ILL BRAINLIST U AND GIVE U 5 STAR TO WHOEVER IS GOOD;) be s

marusya05 [52]

Answer:

A) Oxygen

B) Heat

C)

Explanation:

Hope it helps

4 0

3 years ago

Read 2 more answers

What products result from mixing aqueous solutions of Cr(NO3)2(aq) and NaOH(aq)? Question 10 options: Cr(OH)2(s), Na+(aq), and N

Alecsey [184]

Answer:

Cr(OH)2(s), Na+(aq), and NO3−(aq)

Explanation:

Let is consider the molecular equation;

2NaOH(aq) + Cr(NO3)2(aq) -----> 2NaNO3(aq) + Cr(OH)2(s)

This is a double displacement or double replacement reaction. The reacting species exchange their partners. We can see here that both the sodium ion and chromium II ion both exchanged partners and picked up each others partners in the product.

Sodium ions and nitrate ions now remain in the solution while chromium II hydroxide which is insoluble is precipitated out of the solution as a solid hence the answer.

4 0

3 years ago