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podryga [215]
3 years ago
13

A NaOH(aq) solution with a pH value of 13 is used to determine the molarity of a HCl(aq) solution. A 10.0-mL sample of the HCl(a

q) is exactly neutralized by 16.0 mL of 0.100 M NaOH(aq). During this laboratory activity, appropriate safety equipment was used and safety procedures were followed. Determine the pH value of a solution that has a H+(aq) ion concentration 10 times greater than the original NaOH(aq) solution.
Chemistry
1 answer:
erica [24]3 years ago
5 0

Answer:

pH of the H⁺(aq) is 0

Explanation:

It is possible to know the  concentration of a HCl(aq) solution by titration with a solution of NaOH(aq) with known concentration. The reaction is:

HCl(aq) +  NaOH(aq) → H₂O(l) + NaCl(aq)

The added moles of NaOH are equal to moles of HCl and as you know volume of HCl added you will obtain concentration of HCl.

Now, a solution of H⁺(aq) with a concentration 10 times greater than original NaOH(aq) solution -0.100M-, has a concentration of 1.00M H⁺(aq), the pH of this solution is:

pH = -log (1.00M H⁺(aq) = 0

That means <em>pH of the H⁺(aq) is 0</em>

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Answer:

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b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

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Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

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Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

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3 years ago
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Answer:

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