Post by: username on October 14, 2008, 08:18:34 pm
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! :)
Post by: Mao on October 14, 2008, 08:39:29 pm
2. lets shorthand "hypobromous acid" to HB, and its conjugate ion B-
in this case,
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
Post by: username on October 14, 2008, 09:29:52 pm
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!
Post by: Mao on October 14, 2008, 09:52:21 pm
4. and silly me. the equation is
Post by: username on October 21, 2008, 05:33:57 pm
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?
Post by: Mao on October 21, 2008, 05:39:47 pm
Post by: username on October 21, 2008, 05:51:46 pm