First, we need to calculate moles of hydrazoic acid NH3:
moles NH3 = molarity * volume
= 0.15 m * 0.025 L
= 0.00375 moles
moles NaOH = molarity * volume
= 0.15 m * 0.015 L
= 0.00225 moles
after that we shoul get the total volume = 0.025L + 0.015L
= 0.04 L
So we can get the concentration of NH3 & NaOH by:
∴[NH3] = moles NH3 / total volume
= 0.00375 moles / 0.04 L
= 0.09375 M
∴[NaOH] = moles NaOH / total volume
= 0.00225 moles / 0.04 L
= 0.05625 M
then, when we have the value of Ka of NH3 so we can get the Pka value from:
Pka = -㏒Ka
= - ㏒ 1.9 x10^-5
= 4.7
finally, by using H-H equation we can get PH:
PH = Pka + ㏒[salt/ basic]
PH = 4.7 +㏒[0.05625/0.09375]
∴ PH = 4.48
An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.
Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]
moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles
moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles
pKa of phenol = 9.98
We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation
pH = pKa + log [salt] / [acid]
volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula
pH = 9.98 + log [12.93 / 42.55] = 9.46
Answer:
Thus, the order of the reaction is 2.
The rate constant of the graph which is :- 2.00 M⁻¹s⁻¹
Explanation:
The kinetics of a reaction can be known graphically by plotting the concentration vs time experimental data on a sheet of graph.
The concentration vs time graph of zero order reactions is linear with negative slope.
The concentration vs time graph for a first order reactions is a exponential curve. For first order kinetics the graph between the natural logarithm of the concentration vs time comes out to be a straight graph with negative slope.
The concentration vs time graph for a second order reaction is a hyberbolic curve. Also, for second order kinetics the graph between the reciprocal of the concentration vs time comes out to be a straight graph with positive slope.
Considering the question,
A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M⁻¹s⁻¹.
<u>Thus, the order of the reaction is 2.</u>
<u>Also, slope is the rate constant of the graph which is :- 2.00 M⁻¹s⁻¹</u>
Answer:
Supersaturated solution.
Explanation:
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In this case, according to the types of solution in terms of the relative amounts of solute and solvent, we can define a point called solubility at which the amount of solute is no longer dissolved in the solvent; thus, a value of solute/solvent less than the solubility is related to unsaturated solutions, equal to the solubility is related to the saturated solutions and more than the solubility to supersaturated solutions.
Thus, since solubility is temperature-dependent, at 30 °C the solubility of sodium chloride is 36.09 g per 100 mL of water; which means that, since the solution has 50 g of sodium chloride, more than 36.09 g, we infer this is a supersaturated solution.
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