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3 years ago

What is the ph of 0.450 m al(no3)3 [ka for al3+(aq) = 1.00x10-5]? express your answer to two decimal places?

Chemistry

1 answer:

White raven [17]3 years ago

8 0

Answer : The pH of the solution is, 2.67

Explanation :

The equilibrium chemical reaction is:

$Al^{3+}+H_2O\rightarrow Al(OH)^{2+}+H^+$

Initial conc. 0.450 0 0

At eqm. (0.450-x) x x

As we are given:

$K_a=1.00\times 10^{-5}$

The expression for equilibrium constant is:

$K_a=\frac{(x)\times (x)}{(0.450-x)}$

Now put all the given values in this expression, we get:

$1.00\times 10^{-5}=\frac{(x)\times (x)}{(0.450-x)}$

$x=0.00212M$

The concentration of $H^+$ = x = 0.00212 M

Now we have to calculate the pH of solution.

$pH=-\log [H^+]$

$pH=-\log (0.00212)$

$pH=2.67$

Therefore, the pH of the solution is, 2.67

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A piece of high-density Styrofoam measuring 24.0 cm by 36.0 cm by 5.0 cm floats when placed in a tub of water. When a 1.5 kg boo

zheka24 [161]

Explanation:

The given data is as follows.

Width of Styrofoam = 24.0 cm

Length of Styrofoam = 36.0 cm

Height of Styrofoam = 5.0 cm

Therefore, volume of the Styrofoam will be calculated as follows.

Volume = length × width × height

= (36.0 × 24.0 × 5.0) $cm^{3}$

= 4320 $cm^{3}$

or, = $4.32 \times 10^{3} cm^{3}$

As Styrofoam partially sinks at 3.0 cm and total height of Styrofoam is 5.0 cm. Hence, height of Styrofoam above the water is (5.0 - 3 cm) = 2 cm.

So, volume of water displaced is as follows.

24.0 cm × 36.0 cm × 2.0 cm

= $1.73 \times 10^{3} cm^{3}$

Hence, mass of displaced water is as follows.

mass = density × volume

= $1.00 g/cm^{3} \times 1.73 \times 10^{3} cm^{3}$

= $1.73 \times 10^{3} g$

Since, book is placed on the Styrofoam. Therefore, mass of water displaced is also equal to the following.

Mass of water displaced = mass of book + mass of Styrofoam

$1.73 \times 10^{3} g$ = 1500 g + mass of Styrofoam

(1730 - 1500) g = mass of Styrofoam

mass of Styrofoam = 230 g

Therefore, calculate the density of Styrofoam as follows.

Density = $\frac{mass}{volume}$

= $\frac{230}{4.32 \times 10^{3} cm^{3}}$

= $53.24 \times 10^{-3} g cm^{-3}$

Thus, we can conclude that the density of Styrofoam is $53.24 \times 10^{-3} g cm^{-3}$ .

4 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

15 grams of sodium reacts with a certain amount of chlorine to yield 37 grams of sodium chloride, as shown below.

ale4655 [162]

Answer: 22g of chlorine would be needed to carry out this synthesis reaction

Explanation:

A synthesis reaction is one in which two or more than two elements combine together to forma single product.

$Na+Cl\rightarrow NaCl$

The atoms present in the reactants are found on the product side. According to the law of conservation of mass, the number of atoms on both sides of the arrow must be same as the total mass must be conserved.

15 grams of sodium reacts with 22 grams of chlorine to yield 37 grams of sodium chloride. Thus 22g of chlorine would be needed to carry out this synthesis reaction.

4 0

3 years ago

1.34 milligrams is the same as _____ kg and _____g.

Karolina [17]

Answer C is for kg and but it's .00134 for grams

3 0

3 years ago

Read 2 more answers

How do you find the molar mass of a hydrate

Darya [45]

If you mean hydrate as in <em>MgSO4 · 7H2O, </em>then simply find the molar mass of each element you see.

For the example above, that means you would add the molar mass (found on the periodic table) of Mg, then S, then 4(O), 14(H), and 7(O).

The results would be your molar mass for the hydrate.

I hope this is what you meant by your question!

6 0

3 years ago