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Ilya [14]
2 years ago
9

For which salt is the aqueous solubility most likely to change by changing the pH of the solution?

Chemistry
1 answer:
Hoochie [10]2 years ago
4 0

The salt whose aqueous solubility will most likely to change the pH of the solution is CaF2(aq).

The pH of a salt solution depends on the nature of the acid and base that forms the salt.

  • Salt of a weak acid and a strong base yields a basic solution
  • Salt of a strong acid and a weak base yields an acidic solution
  • Salt of a strong acid and a strong base yields a neutral solution.

Since HF is a weak acid, its salt CaF2 will be slightly basic when dissolved in water hence the pH of the solution is altered.

Learn more: brainly.com/question/14072179

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You have 125.0 mL of a solution of H3PO4, bu you don't know its concentration. if you titrate the solution with a 4.56-M solutio
irakobra [83]

Answer:

4.90 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = ? M

V₁ = 125.0 mL

M₂ = 4.56 M

V₂ = 134.1 mL

Using the above formula , the molarity of acid , can be calculated as ,

M₁V₁ = M₂V₂  

Substituting the respective values ,  

M₁ *  125.0 mL = 4.56 M *  134.1 mL

M₁ = 4.90 M

6 0
3 years ago
A solution has a [OH−] of 1 x 10−12. What is the pOH of this solution?<br> 2<br> 7<br> 10<br> 12
strojnjashka [21]
-log (1×10^-12) is how you calculate the pOH which in this case is 12
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Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,
8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0
3 years ago
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