A chemist dissolves 699. mg of pure hydrobromic acid in enough water to make up 180. mL of solution. Calculate the pH of the sol
ution. Be sure your answer has the correct number of significant digits.
1 answer:
Answer: 1.32
Explanation:
First, we must obtain the molar mass of HBr. After that, we try to obtain the concentration of the hydrobromic acid from the formula n=CV since the volume of solution and mass of acid was provided. Recall that n=m/M. If the concentration of acid is thus obtained, we make use of the fact that the concentration of H+ in the acid is equal to the molar concentration of HBr to obtain the pH. The pH is the negative logarithm of the concentration we obtained in the initial step.
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Answer : The partial pressure of at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar
Solution : Given,
Initial pressure of = 1.42 bar
Initial pressure of = 2.87 bar
= 0.036
The given equilibrium reaction is,
Initially 1.42 2.87 0
At equilibrium (1.42-x) (2.87-3x) 2x
The expression of will be,
Now put all the values of partial pressure, we get
By solving the term x, we get
From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.
Thus, the partial pressure of at equilibrium = 2x = 2 × 0.287 = 0.574 bar
The partial pressure of at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar
The partial pressure of at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar
The total pressure at equilibrium = Partial pressure of + Partial pressure of
+ Partial pressure of
The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar