An analytical chemist is titrating of a solution of hydrazoic acid with a solution of . The of hydrazoic acid is . Calculate the
pH of the acid solution after the chemist has added of the solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of solution added. Round your answer to decimal places.
1 answer:
Answer:
ph= 3.53
Explanation:
A pH of 7 is neutral. A pH less than 7 is acidic. A pH greater than 7 is basic. The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value.
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Charles law gives the relationship between temperature of gas and volume of gas.
It states that for a fixed amount of gas, temperature is directly proportional to volume of gas.
V / T = k
where V- volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
T1 = 250 °C + 273 = 523 K
T2 = 150 °C + 273 = 423 K
Substituting the values in the equation,
V = 251 mL
the new volume is 251 mL
It states that for a fixed amount of gas, temperature is directly proportional to volume of gas.
V / T = k
where V- volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
T1 = 250 °C + 273 = 523 K
T2 = 150 °C + 273 = 423 K
Substituting the values in the equation,
V = 251 mL
the new volume is 251 mL
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e.
Calculations:
The second order rate equation is:
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M