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brilliants [131]
3 years ago
10

By how much will the ph change if 0.025 mol of hcl is added to 1.00 l of the buffer that contains 0.15 m hc2h3o2 and 0.25 m c2h3

o2–? use ka = 1.8 × 10–5 for hc2h3o2.
Chemistry
1 answer:
Setler [38]3 years ago
3 0
According to the reaction equation:

              CH3COO-  + H+   → CH3COOH 

initial         0.25                             0.15

change     - 0.025                          + 0.025

Equ         (0.25-0.025)                 (0.15 + 0.025)

first, we have to get moles acetate and moles acetic acid:

moles of acetate = 0.25 - 0.025 = 0.225 moles

∴ [CH3COO-] = 0.225 mol / 1 L = 0.225 M

moles of acetic acid = 0.15 + 0.025 = 0.175 moles

∴ [ CH3COOH] = 0.175 mol / 1L = 0.175 M 

Pka = -㏒ Ka 

       = -㏒ 1.8 x 10^-5
       
       = 4.74

from H-H equation we can get the PH value:

PH = Pka + ㏒ [acetate / acetic acid]

PH = 4.74 + ㏒[0.225/0.175]

∴ PH = 4.8
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Answer:

1.  the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c (\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} })

Explanation:

To find  -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

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Before filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 117 kPa

⇒Absolute pressure ( p1 )  = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 212 kPa

⇒Absolute pressure (p2)  = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. V_{1} = V_{2}

As we know that, P ∝ M

⇒ \frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }

⇒\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }

⇒\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378 ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒\frac{P}{T} ∝ M

⇒\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }

So, The correct relation is c option.

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