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serg [7]
3 years ago
6

. One likely reason for the decreased effectiveness of this chemical is that these bacteria have

Biology
1 answer:
wel3 years ago
3 0
A mutation for resistance
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Be sure to answer all parts.
Aleks04 [339]

Answer:

pH = 3.215

Explanation:

From the given information;

Using the equation for the dilution of  a stock solution:

Since moles of C5H5N = moles of HNO3

Then:

M_{C_5H_5N}\times V_{C_5H_5N}= M_{HNO_3}\times V_{HNO_3}

0.0750 M \times 2.65 L = 0.407 M \times V_{HNO_3}

V_{HNO_3}= \dfrac{0.0750 M \times 2.65 L}{0.407 M}

V_{HNO_3}=488.33 \ mL

The reaction between C5H5NH and H2O is as follows:

C_5H_5N^+H + H_2O ⇄ C_5H_5N + H_3O^+

Molarity  \ of \ C_5H_5N^+H = \dfrac{moles}{addition \ of\  the\  total\ volume}

Molarity  \ of \ C_5H_5N^+H = \dfrac{0.0750 \ M \times 2.65 \ L}{2.65 \ L + 0.48833 \ L}

Molarity  \ of \ C_5H_5N^+H = \dfrac{0.19875 \ ML}{3.13833 L }

= \ 0.06333\ M

Now, the next step is to draw out the I.C.E table.

    C_5H_5N^+H + H_2O ⇄ C_5H_5N + H_3O^+

I       0.06333                      0          0

C        - x                               x          x  

E      0.06333 -x                   x          x

K_a  = \dfrac{10^{-14}}{1.7 \times 10^{-19}} \\ \\  K_a = 5.8824 \times 10^{-6}

K_a = \dfrac{[C_5H_5N][H_3O^+] }{[C_5H_5N^+H] }  \\ \\  5.8824 \times 10^{-6} = \dfrac{x^2}{0.06333 - x}

Assuming x < 0.06333

x^2 = 5.8824 \times 10^{-6} \times 0.06333

Then

x^2 = 3.72532392 \times 10^{-7} \\ \\ x= \sqrt{3.72532392 \times 10^{-7}} \\ \\  x = 6.1035 \times 10^{-4} \ M

[H_3O^+] =  x = 6.1035 \times 10^{-4} \ M  \\ \\  pH = -log (6.1035 \times 10^{-4}) \\ \\ \mathbf{\\ pH = 3.215}

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