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Marrrta [24]
3 years ago
7

What mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 giv

en that the Ka of C6H5COOH is 6.5×10-5 and the molar mass of NaC6H5COO is 144.1032 g/mol?
Chemistry
1 answer:
statuscvo [17]3 years ago
7 0

Answer:

41 g

Explanation:

We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]

log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40

[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

We can find the mass of NaC₆H₅COO using the following expression.

M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L

mass NaC₆H₅COO = 41 g

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Dimas [21]
<h2>Hello!</h2>

The answer is:

The new volume is equal to 206.5 L.

<h2>Why?</h2>

To solve this problem, we need to assume that the pressure is constant, and use the Charle's Law equation, so, solving we have:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

We are given:

V_1=350L\\T_1=500K\\T_2=295K

Then, using the Charle's Law equation, we have:

\frac{350L}{500K}=\frac{V_2}{295K}

\frac{350L}{500K}=\frac{V_2}{295K}\\\\V_2=\frac{350L}{500K}*295K=206.5L

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Have a nice day!

5 0
2 years ago
How many moles of sulfur<br> dioxide are in 2.26 x 10^33 sulfur dioxide molecules?
LUCKY_DIMON [66]

Answer:

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         = 2.26 x 10^33 / 6.022 x 10^23

         = 3752906011

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6 0
3 years ago
A simple machine produces 25 joules of output work for every 50 joules of input work . What is the efficiency of this machine ?
nataly862011 [7]
2:1, output is half the input
8 0
3 years ago
Please answer, with explanation. Thanks!​
nadya68 [22]

Answer:

Explanation:

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Number of moles = 2.65 g/ 55.85 g/mol

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6 0
3 years ago
Consider the following system at equilibrium where H° = 111 kJ/mol, and Kc = 6.30, at 723 K.
Rashid [163]

Answer:

1) The value of Kc:

C. remains the same.

2) The value of Qc:

A. is greater than Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium.

4) The concentration of N2 will:

B. decrease.

Explanation:

Hello,

In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:

1) The value of Kc:

C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

ln(K)=\frac{\Delta _RG}{RT}

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A. is greater than Kc, since the reaction quotient is:

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3) The reaction must:

B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.

4) The concentration of N2 will:

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Best regards.

8 0
3 years ago
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