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inn [45]
4 years ago
5

What is the pH of a 900 mL solution containing 3.40 grams of hydrocyanic acid

Chemistry
1 answer:
Allushta [10]4 years ago
7 0

Answer:

pH = 5.05

Explanation:

pH is derived from the concentration of hydronium ions in a solution. Hydrocyanic acid is HCN.

First, we shall figure out the moles of HCN:

\frac{3.4g}{27.03g/mol}  = 0.125786

If HCN was a strong acid:

HCN has a 1:1 ratio of H+ ions, the moles of H+ is also the same.

To find the molarity, we now divide by Liters. This gets us:

\frac{0.125786 moles}{0.9L} = 0.139762 M

Finally, we plug it into the definition of pH:

pH = -log[H^{+} ]

pH = -log(0.139762)

pH = 0.855

However, since HCN is a weak acid, it only partially dissociates. The K_a of HCN is 6.2 * 10^{-10}.

K_a = \frac{[H^+][A^-]}{[HA]}

We can use an ice table to determine that when x = H+,

K_a = \frac{x^2}{0.125786-x}

[H^+] = 8.83*10^{-6}

pH = -log[H^{+} ]

pH = -log(8.83 * 10^{-6} )

pH = 5.05

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The specific heat is  c_b  = 0.402 J / g \cdot ^oC

Explanation:

From the question we are told that

    The mass of  the sample is  m = 54.4 \ g

     The mass of the water is  m_w = 150.0 \ g

     The initial temperature of the sample is  T_i  = 95.1 ^oC

     The initial temperature of the water is  T_{w_i} =  15^oC

     The final temperature of the water is  T  =  17.6 ^oC

Note the final temperature of water is equal to the final temperature of brass sample

    The pressure is  P =1 \ atm

Generally for according to the law of energy conservation

    The heat lost by sample  =  The heat gain by water

   

The heat lost by brass sample is  mathematically evaluated as    

          H_L  =  m * c_b  *  [T_i - T]

Where c_b is the specific neat of the brass sample

The heat gained  by water is  mathematically evaluated as          

        H_g  = m_w *c_w * [T_w - T ]

where c_w is the specific heat of water which has a constant value of  

     c_w =  4.186  joule/gram

So

    H_L  =  H_g \ \equiv m* c_b  * [T_i -T] =  m_w * c_w * [T - T_w]

substituting values

    52.4 * c_b  * [95.1  - 17.6] =  150 * 4.186 * [ 17.6 - 15.0]

    c_b  = 0.402 J / g \cdot ^oC

4 0
3 years ago
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