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Lady bird [3.3K]
3 years ago
12

A solution containing potassium bromide is mixed with one containing lead acetate to form a solution that is 0.013 M in KBr and

0.0035 M in Pb(C2H3O2)2.Will a precipitate form in the mixed solution? If so, identify the precipitate.Express answer as a chemical formula.Please explained how you determined the answer.

Chemistry
1 answer:
galben [10]3 years ago
7 0

Answer:

No precipitate will form on mixing potassium bromide and lead acetate

Explanation:

Precipitation occurs when the ionic product of the ions exceeds the solubility product of the corresponding salt in the solution. If, this implies that the solution is over saturated and precipitation of salt will occur.Potassium bromide and lead acetate are strong electrolyte and thus completely dissociate in water.

The dissociation reaction is given as,Potassium bromide and lead acetate react to give lead bromide and potassium acetate.The concentration of lead ions is equal to the concentration of.The concentration of bromide ion is equal to the concentration of.Potassium acetate is soluble in aqueous solution. Lead bromide will precipitate out if the ionic product of lead ion and bromide ion exceeds the solubility product of .

The value of solubility product constant,, for is.The reaction quotient for is calculated as,Substitute the values.The value of the reaction quotient for is less than its solubility product, , therefore, precipitation of will not occur.

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How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m
Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

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

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol

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

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

pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})

We are given:

pK_a = negative logarithm of acid dissociation constant of acetic acid = 4.74

[CH_3COONa]=?mol  

[CH_3COOH]=0.200mol

pH = 5.00

Putting values in above equation, we get:

5=4.74+\log(\frac{[CH_3COONa]}{0.200})

[CH_3COONa]=0.364mol

To calculate the mass of sodium acetate for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g

Hence, the mass of sodium acetate that must be added is 30.23 grams

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3 years ago
Select all that apply.
lutik1710 [3]
Correct answer is <span>X = ΔH

Reason:
1) The graph of enthalpy Vs reaction coordinate  suggest the reaction is endothermic in nature. For endothermic reaction, energy if product is more than that of reactant. Hence,  option 1 i.e. </span><span>X = -ΔH cannot be correct.
2) Since the reaction is endothermic in nature, </span>energy if product is more than that of reactant. Hence,  option 2 i.e. X = ΔH is correct.
3) Activation energy is energy difference between Reactant (A) and transition state (B). However, as per option C, activation energy (A.E.) is energy difference between product (C) and transition state (B), which is incorrect. 
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