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Papessa [141]
3 years ago
11

Calculate the concentration of buffer components present in 295.00 mL of a buffer solution that contains 0.390 M NH4Cl and 0.390

M NH3 immediately after the addition of 1.40 mL of 6.00 M HNO3.
Chemistry
1 answer:
aivan3 [116]3 years ago
8 0

Answer:

Explanation:

The acid HNO₃ will react with the base NH₃ in the buffer to produce NH₄⁺, changing the concentrations of both NH₃ and NH₄⁺ in the buffer.

Therefore, this question is solved by performing stoichiometric calculations based on those reactions:

HNO₃       +      NH₃     ⇒        NH₄⁺

Vol HNO₃ = 1.40 mL x 1L / 1000 mL = 1.4 x 10⁻³ L

# mol HNO₃ = 1.4 x 10⁻³ L x 6.00 mol/L =  8.4 x 10⁻³ mol HNO₃

Since the reaction occurs in 1:1 ratio, 8.4 x 10⁻³ mol NH₃ will be consumed and 8.4 x 10⁻³ mol of NH₄⁺ will be produced.

Initially we had

# mol NH₄⁺ = 0.295 L x 0.390 mol/L =  0.115 mol

# mol NH₃ = 0.115

After reaction we will have

# mol NH₄⁺  = 0.115 mol + 8.4 x 10⁻³ mol = 0.123 mol

# mol NH₃ = 0.115 mol - 8.4 x 10⁻³ mol = 0.107 mol

Therefore, their new concentrations will be

Volume after reaction = 0.295 L +0 .0014 L = 0.296 L

[NH₄⁺] = 0.123 mol / 0.296 L = 0.416 M

[NH₃] = 0.107 mol / 0.296 L = 0.361 M

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From the question given, we obtained the following data:

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In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o
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Answer:

Stock  solution =  1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

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<u>Step 1:</u> Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

<u>Step 2</u>: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

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<u>Step 3:</u> Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

<u>Step 4</u>: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

<u>Step 5:</u> Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

<u>Step 6</u>: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

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