Help with trial exam questions?

[username]

1. A solution of lime water contains 5 x 10^-5 mol of Ca(OH)2 IN 10ml of the solution at 25 degrees. The pH of the solution is?
What I did was ( 10^-14 / (5.00 x 10^-5 x 2) ) and then -log(answer) it. But it's wrong. The answer should be 2?

2. A 10ml solution of hypobromous acid has pH of 5. What is the concentration of the acid? (The Ka value is 2.4 x 10^-9)

3. (Unrelated to pH but).. Can someone explain this question:
A well insulated calorimeter is first calibrated electrically and the temperature noted. A substance is dissolved and the temperature is again noted. The reaction is endothermic. The temperature noted in the calorimeter rises then falls. I dont understand why it rises and then falls if the reaction is endothermic.. wouldn't it rise again since it should be absorbing heat?

4. An excess of zinc powder was added to 50ml of .1M AgNO3 in a polystyrene cup. Initially the temperature of the solution was 21.10 degrees and during the reaction it rose to 25.40 degrees. EDIT: Assuming the density of the solution is 1g/ml, and ignoring the heat capacity of the metals, calculate the enthalpy change for the reaction. (Yeah sorry I completely forgot the rest of the question earlier!)


Once again thanks for any help!

[Mao]

1. look at the chemical formula, it is Ca(OH)₂, i.e. the hydroxide concentration, when dissolved, is twice the given concentration.

2. lets shorthand "hypobromous acid" to HB, and its conjugate ion B-

in this case, and
also, since the pH is high, we can assume it is a weak acid:




3. when you calibrate it electrically, electrical energy is converted to thermal energy in the heating element, and heats up the water. Then, the endothermic reaction takes place, which absorbs thermal energy, and the temperature falls.

4. E=mct is awkwardly expressed. it refers to the specific heat capacity (joules per degree per gram), an indication of how well a particular material retains heat.
1 gram of water takes 4.18 joules of energy to heat it by 1 degree, and the density of water is 1g/mL (i.e. 1mL weighs 1g), hence 50mL of water would take 50*4.18 joules of energy to heat it by 1 degree.
in this case, it has been heated 4.3 degrees, so it takes (50*4.18)*4.3 joules of energy in total

assuming you now want the heat of reaction, which is usually in kJ/mol, also notice that the temperature increased, so the figure is negative

[username]

1. For the first question, I still can't get the right answer which should be 2. I doubled the OH concentration, and divided that by 10^-14, but still got 10^-10.

4 They did something differently and got a different answer. They did ..

E = m x c x T
  = 50 x 4.18 x 4.3
  = 899 J

899J / (5.00 x 10^-3) mol = xJ / 2 mol

x = 3.60 x 10^5 J
x = 360 kJ

But I did your answer and weirdly enough if you just multipled by two it'd be near it?

Thanks a lot for 2 and 3!

[Mao]

1. you are right. a pH of 2 would indicate an acid, whereas this is a base. the answer is wrong.

4. and silly me. the equation is , hence to get the heat of reaction for this reaction, you would multiply it by 2.

[username]

Glucose burns in excess oxygen according to

C6H1206 + 6O2 --> 6CO2 + 6H2O

When a 1.324 sample of glucose was burned in the calorimeter with calibration factor 1.18, the temperature increased by 17.32. Calculate the heat of combustion of glucose in kJ per gram.

This is where I've gotten to so far.

/\E = CF x T
     = 1.18 x 17.32 = 20.43

n(glucose) = 1.324/180 = .00735

Heat of combustion = E/n
                           = 20.43/.00735

But that gives me the wrong answer.. How come?

[Mao]

the question asks per gram, so its not necessary to convert it to number of moles.

[username]

Ooh okay thanks. I thought the H formula thingo was strictly limited to mol only. u_u