pH question (1 Viewer)

[stargaze]

hi ppls,

how would it be best go about doing something like this:

Calculate the pH of the solution produced by mixing 50mL of 0.1 mol L^-1 HCl with 20mL of 0.05 mol L^-1 NaOH

Cant seem to find an explanation of these types in a textbook

thanks

 

[Ednaw]

pH is -log base 10 of the moles L_-1 ....hmm come online man lets talk about it il give u the explanation i get from a book

 

[Poon-Tang]

the best way is to get the chemicals and do the experiment, with one of those PH measuring peices of equipment, i dunno what their called i dont do chemistry, but that would be the best way, cos it would be the most accurate, it would be a "1st hand investigation" and therefore must be right...

anyway there is probably a faster way of doing it, but if i knew the ph leverl of .1 molar hydrocloric acid and .05 molar NaOH i could do it could it would be the the average of the two in a ratio of 5:2.

 

[wogboy]

HCl + NaOH -> NaCl + H<SUB>2</SUB>O

You have 0.05*0.1 = 0.005 moles of HCl and 0.02*0.05 = 0.001 moles of NaOH. After the reaction, you'll have 0.004 moles of unreacted HCl remaining.

The concentration of HCl is 0.004 moles / 70 mL = 0.057 mol/L = Concentration of H<SUP>+</SUP>

pH = -log(H<SUP>+</SUP> concentration)
= -log(0.057)
= 1.24

 

[stargaze]

ahh icic .. thanks wogboy

um just another q for anyone... never really understood how to do those ones where u have ethanol heating up water.. e.g.

Mr. X used spirit burners to compare the heats of combustion of ethanol. He burned 0.4g of ethanol using the flame to heat 100mL of water. lTemperature rose from 20 to 34C.

a) Given specific heat of water is 4.2x10^3 J kg^-1 K^-1, calculate the experimental molar heat of combustion of ethanol.

b) The experiment was repeated using methanol. If accurate values for the heats of combustion of methanol and ethanol are 727 and 1367kJ mol^-1 respectively, estimate the value this student would expect to obtain for methanol.

thanks ~

my bad.. yeah, the q was missign some stuff. edited

 

[Dreamerish*~]

ahh icic .. thanks wogboy

um just another q for anyone... never really understood how to do those ones where u have ethanol heating up water.. e.g.

Mr. X used spirit burners to compare the heats of combustion of ethanol, using the flame to heat 100mL of water. Temperature rose from 20 to 34C.

a) Given specific heat of water is 4.2x10^3 blah .. calculate the experimental molar heat combustion of ethanol.

b) The experiment was repeated using methanol. If accurate values for the heats of combustion of methanol and ethanol are 727 and 1367kJ mol^-1 respectively, estimate the value this student would expect to obtain for methanol.

thanks ~

are you sure that the questions were worded like that?
how many grams of ethanol was used in a)?

 

[stargaze]

ahh icic.. sorry about that.. i've edited it..

 

[m_isk]

ahh icic .. thanks wogboy

um just another q for anyone... never really understood how to do those ones where u have ethanol heating up water.. e.g.

Mr. X used spirit burners to compare the heats of combustion of ethanol. He burned 0.4g of ethanol using the flame to heat 100mL of water. lTemperature rose from 20 to 34C.

a) Given specific heat of water is 4.2x10^3 J kg^-1 K^-1, calculate the experimental molar heat of combustion of ethanol.

b) The experiment was repeated using methanol. If accurate values for the heats of combustion of methanol and ethanol are 727 and 1367kJ mol^-1 respectively, estimate the value this student would expect to obtain for methanol.

thanks ~

my bad.. yeah, the q was missign some stuff. edited

(a) using deltaH = -mcdelta T...
delta H =molar heat of substnace x no. of mols of substnace (if you dont believe me, try substituting the units...molar heat (kj/mol) x no of mols (mol) = just kJ, which is the units for H so we're cool )
therefore,
(molar heat) x (0.4/46)= 0.1*4.12*(34-20)
Note: 46 is molar mass of ethanol, and 100ml is 0.1L..also, you should show the units for everything in ur calculation..could get messy typing tho
anyway....by calculation, molar heat is (0.1*4.12*14)/(0.4/46) = 663.32kJ/mol
(really the answer should be negative cos the reaction is exothermic, but the - is commonly left out..cos we normally work out the molar heat of combustion for fuels and duh fuels will release energy (in the form of heat) and hence have a - heat of reaction, and this reaction is combustion in this case!!)

As for b, i reckon if we have a difference of (1367-663)/1367*100 =51.4% error for ethanol, we should have the same for methanol. THerefore, if its heat of combustion is 727kJ/mol, the experimental one should be 51.4% of this, which is 374.4 kJ/mol.

I am 99% sure about me answer for a, but not so for b...could some genius clear it up?? thanks

 

[stargaze]

for b)

my book says this:

727/1367 = x/677

where x is the value for methanol

in the marking criteria it says:

calculates ratio of heats of combustion ..........................(1)
727/1367 = x/677
realises experimental results should be in similar ratio (360 kJ mol-1) .............(1)

now, i dont quite understand so if someone does, could they elaborate please?