Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] - Education (12)

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2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } / Nairaland 2016 Jamb Tutorial Classroom [government Thread] / Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] (2) (3) (4)

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Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Nobody: 10:42pm On Feb 21, 2016

Geofavor:
Calculate the pH of a solution of 0.0001 moldm^-3 hydrochloric acid.

Icecalm, oluchidelly, mathefaro, shyneptune, oxytocin

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Geofavor(m): 10:46pm On Feb 21, 2016

ShyNeptune:
4

show working abeg

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Nobody: 10:53pm On Feb 21, 2016

Geofavor:

show working abeg

Done

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Oxytocin(m): 1:27am On Feb 22, 2016

Geofavor:
Calculate the pH of a solution of 0.0001 moldm^-3 hydrochloric acid.

Icecalm, oluchidelly, mathefaro, shyneptune, oxytocin

pH=-log[H+]

pH=-log 0.0001

pH=-log10^-4

pH=-(-4)

pH= +4

**** Modified***

@ShyNeptune...Good
@Mathefaro...Thanks...Seen the error

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by mathefaro(m): 1:41am On Feb 22, 2016

Oxytocin:

pH=-log[H+]

pH=-log 0.0001

pH=-log10^-5

pH=-(-5)

pH= +5

Please check again, it's 4
@Geofavor, the pH of a solution is defined as the negative logarithm of the hydrogen ion concentration present in that solution to base 10.
I.e pH = -log₁₀[H⁺].

For the question, first and foremost, you need to know how HCL ionizes in water to know the concentration of hydrogen ion present in one mole of the solution
HCL = H⁺ + CL^-
Therefore, it means that 1 mole if HCL contains 1 mole of H⁺, so 0.0001mole of the acid will also contain 0.0001mole of H⁺. The rest is what oxytocin have solved above.

Mind you if it were to be H₂SO₄, it ionizes to give 2 moles of H⁺ since it is dibasic unlike HCL which is monobasic.
Also note that the H⁺ in the formula above can be changed to H₃O⁺ while
pOH = -log₁₀[OH^-].
And pH + pOH = 14

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by serenityaliyu(m): 5:08am On Feb 22, 2016

Geofavor:
Calculate the pH of a solution of 0.0001 moldm^-3 hydrochloric acid.

Icecalm, oluchidelly, mathefaro, shyneptune, oxytocin

0.0001=1×10^-4.
=-log10(1×10^-4)=
-log1-log10^-4=0+4=4.

please quote me if am wrong.

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Geofavor(m): 8:30am On Feb 22, 2016

mathefaro:
Please check again, it's 4
@Geofavor, the pH of a solution is defined as the negative logarithm of the hydrogen ion concentration present in that solution to base 10.
I.e pH = -log₁₀[H⁺].

For the question, first and foremost, you need to know how HCL ionizes in water to know the concentration of hydrogen ion present in one mole of the solution
HCL = H⁺ + CL^-
Therefore, it means that 1 mole if HCL contains 1 mole of H⁺, so 0.0001mole of the acid will also contain 0.0001mole of H⁺. The rest is what oxytocin have solved above.

Mind you if it were to be H₂SO₄, it ionizes to give 2 moles of H⁺ since it is dibasic unlike HCL which is monobasic.
Also note that the H⁺ in the formula above can be changed to H₃O⁺ while
pOH = -log₁₀[OH^-].
And pH + pOH = 14

thanks so much for this. I haven't had time to read it and a few other topics too.

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Geofavor(m): 8:34am On Feb 22, 2016

thanks guys.

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Orezy5(m): 11:41am On Feb 22, 2016

200cm³ each of 0.1M solutions of lead(II)trioxonitrate(V) and hydrochloric acid were mixed. Assuming that lead(II)chloride is completely insoluble, calculate the mass of lead(II)chloride that will be precipitated. [Pb=207, Cl=35.5, N=14, O=16]

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Oxytocin(m): 10:46pm On Feb 22, 2016

Orezy5:
200cm³ each of 0.1M solutions of lead(II)trioxonitrate(V) and hydrochloric acid were mixed. Assuming that lead(II)chloride is completely insoluble, calculate the mass of lead(II)chloride that will be precipitated. [Pb=207, Cl=35.5, N=14, O=16]

Are you with options??
Got 139g

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Orezy5(m): 5:27am On Feb 23, 2016

Oxytocin:
Are you with options?? Got 139g

Hmm. U are very far from the correct answer. The answer here is 2.78g

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by OluchiDelly: 5:45am On Feb 23, 2016

watching in 3d cool

interesting to see the young profs doing justice to the right course.
In addition to Mathefaro's, if the basicity is 2 (or 3) , multiply the hydrogen ion conc given by 2 (or 3) respectively before you apply the formula.

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Nobody: 7:21am On Feb 23, 2016

pls solve DAT queation cool

pls solve DAT queation

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Oxytocin(m): 8:04am On Feb 23, 2016

Orezy5:

Hmm. U are very far from the correct answer. The answer here is 2.78g

Hmmm....
C.c...ThankyouJesus, Mathefaro, shyneptune, Geofavor

Did it again and got 5.56g.. Don't know what am not doing right.

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by OluchiDelly: 9:28am On Feb 23, 2016

Oxytocin:

Hmmm....
C.c...ThankyouJesus, Mathefaro, shyneptune, Geofavor

Did it again and got 5.56g.. Don't know what am not doing right.

I doubt Orezy5 answer. Let him tell show us the workings of the author.

This is mine.

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Orezy5(m): 10:24am On Feb 23, 2016

OluchiDelly:

I doubt Orezy5 answer. Let him tell show us the workings of the author.

This is mine.

The question is from lamlad.
Their solution:

The reaction involved is:

2HCl_(aq) + Pb(NO₃)₂_(aq)---->PbCl₂_(s) + 2HNO₃_(aq)

At the start:

n_HCL=(200 x 0.1)/1000= 0.02mol

nPb(NO3)₂=(200 x 0.1)/1000=0.02mol

from the equation,

If 2moles HCl--->1mole Pb(NO3)2

0.02mole HCl--->0.01mole Pb(NO3)2

Since the available mole of Pb(NO3)2 is greater than what is required stoichiometrically, HCl is the limiting reagent, that is, HCl determines the amount of PbCl2 formed.

Also from the equation,

2moles HCl form 1mole PbCl₂

0.02mole HCl will form 0.01mole PbCl₂

PbCl₂= 207 + (35.5 x 2)=278g/mol

mass of PbCl2 precipitated= 0.01mol x 278g/mol

=2.78g

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Nobody: 10:41am On Feb 23, 2016

Orezy5:

The question is from lamlad.
Their solution:

The reaction involved is:

2HCl_(aq) + Pb(NO₃)₂_(aq)---->PbCl₂_(s) + 2HNO₃_(aq)

At the start:

n_HCL=(200 x 0.1)/1000= 0.02mol

nPb(NO3)₂=(200 x 0.1)/1000=0.02mol

from the equation,

If 2moles HCl--->1mole Pb(NO3)2

0.02mole HCl--->0.01mole Pb(NO3)2

Since the available mole of Pb(NO3)2 is greater than what is required stoichiometrically, HCl is the limiting reagent, that is, HCl determines the amount of PbCl2 formed.

Also from the equation,

2moles HCl form 1mole PbCl₂

0.02mole HCl will form 0.01mole PbCl₂

PbCl₂= 207 + (35.5 x 2)=278g/mol

mass of PbCl2 precipitated= 0.01mol x 278g/mol

=2.78g

Cool

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Orezy5(m): 11:39am On Feb 23, 2016

ShyNeptune:
Cool

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by OluchiDelly: 1:52pm On Feb 23, 2016

[/b]
Orezy5:

HCl is the limiting reagent, that is, HCl determines the amount of PbCl2 formed.
[b]

Agreed. Based on the bolded premise, that was what we forgot to consider . Thanks for sharing.

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Orezy5(m): 2:07pm On Feb 23, 2016

OluchiDelly:
[/b][b]

Agreed. Based on the bolded premise, that was what we forgot to consider . Thanks for sharing.

You're welcome

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Nobody: 4:26pm On Feb 23, 2016

peacebeezy:
good day everyone. good news to all students sitting for the 2016 jamb. call mr Benjamin for jamb answers and guidance on how to make good grades in your exam. call Mr Benjamin on 09057518272 good luck.

sir. I'll be writing on the 3rd of march.... I really need your answers .. Plz help me sir.

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Ajoboss(m): 4:38pm On Feb 23, 2016

ShyNeptune:

sir. I'll be writing on the 3rd of march.... I really need your answers .. Plz help me sir.

guy dnt fall victim 2 dis,its a scam

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Orezy5(m): 5:22pm On Feb 23, 2016

Ajoboss:
guy dnt fall victim 2 dis,its a scam

Lolz. He knows that it's scam

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Ajoboss(m): 5:30pm On Feb 23, 2016

Orezy5:
Lolz. He knows that it's scam

gud 2 knw dat he knows dat

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Oxytocin(m): 6:03pm On Feb 23, 2016

ShyNeptune:

sir. I'll be writing on the 3rd of march.... I really need your answers .. Plz help me sir.

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Oxytocin(m): 6:05pm On Feb 23, 2016

Orezy5:

The question is from lamlad.
Their solution:

The reaction involved is:

2HCl_(aq) + Pb(NO₃)₂_(aq)---->PbCl₂_(s) + 2HNO₃_(aq)

At the start:

n_HCL=(200 x 0.1)/1000= 0.02mol

nPb(NO3)₂=(200 x 0.1)/1000=0.02mol

from the equation,

If 2moles HCl--->1mole Pb(NO3)2

0.02mole HCl--->0.01mole Pb(NO3)2

Since the available mole of Pb(NO3)2 is greater than what is required stoichiometrically, HCl is the limiting reagent, that is, HCl determines the amount of PbCl2 formed.

Also from the equation,

2moles HCl form 1mole PbCl₂

0.02mole HCl will form 0.01mole PbCl₂

PbCl₂= 207 + (35.5 x 2)=278g/mol

mass of PbCl2 precipitated= 0.01mol x 278g/mol

=2.78g

Ohh.. Limiting reagent
Thanks. wink

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Orezy5(m): 6:40pm On Feb 23, 2016

Oxytocin:

Ohh.. Limiting reagent Thanks.

You're welcm sir

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Nobody: 6:41pm On Feb 23, 2016

Oxytocin:

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Nobody: 6:42pm On Feb 23, 2016

Ajoboss:
guy dnt fall victim 2 dis,its a scam

thanks boss. wink

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Ajoboss(m): 7:05pm On Feb 23, 2016

ShyNeptune:
thanks boss.

you welcum sir

Re: Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] by Nobody: 7:35pm On Feb 23, 2016

Geofavor:
Calculate the pH of a solution of 0.0001 moldm^-3 hydrochloric acid.

Icecalm, oluchidelly, mathefaro, shyneptune, oxytocin

sorry, I didn't get the mention earlier.

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