Gchem questions.. help please

[_veo_]

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Practice your chem and help me.. please😕

How many moles of hydrochloric acid must be added to
neutralize a one-liter, saturated solution of X(OH)2, if
the Ksp of X(OH)2 is 4 x 10–18?
A. 1 x 10–6 moles
B. 1 x 10–7 moles
C. 2 x 10–6 moles
D. 2 x 10–7 moles

When the following reaction is balanced, what is the
sum of the ionic charges on the right side of the
equation?
H+ + MnO4– + Fe2+ → Mn2+ + Fe3+ + H2O
A. 5+
B. 7+
C. 10+
D. 17+
E. 0

A calcium hydroxide solution is used in an acid/base
titration. If the solution is 7.69 x 10–3 M Ca(OH)2(aq),
what is its normality?
A. 3.845 x 10–3 N
B. 7.69 x 10–3 N
C. 1.54 x 10–2 N
D. 7.69 x 10–2 N
E. 1.54 x 10–1 N

 

[GRAD]

Practice your chem and help me.. please😕

How many moles of hydrochloric acid must be added to
neutralize a one-liter, saturated solution of X(OH)2, if
the Ksp of X(OH)2 is 4 x 10–18?
A. 1 x 10–6 moles
B. 1 x 10–7 moles
C. 2 x 10–6 moles
D. 2 x 10–7 moles

When the following reaction is balanced, what is the
sum of the ionic charges on the right side of the
equation?
H+ + MnO4– + Fe2+ → Mn2+ + Fe3+ + H2O
A. 5+
B. 7+
C. 10+
D. 17+
E. 0

A calcium hydroxide solution is used in an acid/base
titration. If the solution is 7.69 x 10–3 M Ca(OH)2(aq),
what is its normality?
A. 3.845 x 10–3 N
B. 7.69 x 10–3 N
C. 1.54 x 10–2 N
D. 7.69 x 10–2 N
E. 1.54 x 10–1 N

For # 2, have to do half reactions, for #3, multiply value by 2 to get c

 

[_veo_]

For # 2, have to do half reactions, for #3, multiply value by 2 to get c

Thank you.. Could you explain a bit more why in #3 I have to multiply times 2?

 

[GRAD]

Practice your chem and help me.. please😕

How many moles of hydrochloric acid must be added to
neutralize a one-liter, saturated solution of X(OH)2, if
the Ksp of X(OH)2 is 4 x 10–18?
A. 1 x 10–6 moles
B. 1 x 10–7 moles
C. 2 x 10–6 moles
D. 2 x 10–7 moles

When the following reaction is balanced, what is the
sum of the ionic charges on the right side of the
equation?
H+ + MnO4– + Fe2+ → Mn2+ + Fe3+ + H2O
A. 5+
B. 7+
C. 10+
D. 17+
E. 0

A calcium hydroxide solution is used in an acid/base
titration. If the solution is 7.69 x 10–3 M Ca(OH)2(aq),
what is its normality?
A. 3.845 x 10–3 N
B. 7.69 x 10–3 N
C. 1.54 x 10–2 N
D. 7.69 x 10–2 N
E. 1.54 x 10–1 N

for the first one, you gotta know the molar conc versus the ksp rule....for 3 ions its 4x^3 = ksp

so you get the answer to be 1*10^-6, A

I think its on another thread, the moles vs ksp is given - try looking up molar solubility....just memorize them, thats what i did

 

[fancymylotus]

I think the last one is C,but i could be very wrong.

 

[fancymylotus]

Thank you.. Could you explain a bit more why in #3 I have to multiply times 2?

because CaOH2 donates two OH's...which means its 2N,and the M of the solution is given, so you multiply by 2 to take into account its N.

 

[GRAD]

right, the normality is how many oh's there are or how many h's...remember?

 

[Manyak222]

Hey amsie, shouldnt the first one be 2 *10-6 since x is 1*10^-6 but the concentration of OH is 2x??

 

[GRAD]

no cause its based on molar solubility and you have 3 ions, OH counted for....so you get 4x^3 = Ksp

 

[_veo_]

I think the last one is C,but i could be very wrong.

It is C

 

[fancymylotus]

anything with the general formula MX2 gives you a solubility thingie of 4X3. Point in case: CaOH2--->Ca+2 + 2OH-

[Ca] 2[OH] squared = Ksp. If we say that conc=X, then

X 2 X ^2 = Ksp

therefore:

4X^3 =Ksp

 

[GRAD]

ok yeah cause its hcl, you multiply by 2 i guess....weird ?

 

[_veo_]

for the first one, you gotta know the molar conc versus the ksp rule....for 3 ions its 4x^3 = ksp

so you get the answer to be 1*10^-6, A

I think its on another thread, the moles vs ksp is given - try looking up molar solubility....just memorize them, thats what i did

hmm.. the first one says it should be C

 

[fancymylotus]

hmm.. the first one says it should be C

You have to multiply the X value times two because there are 2OH ions floating around. That will give you answer choice C.

 

[Manyak222]

right ya I know that part. I ended up with x equals 1 *10^-6. But what confuses me is why that is the amount of HCl you have to add, when the concentration of OH- ions is twice that (2x). I guess Im missing something here.😡

 

[fancymylotus]

right ya I know that part. I ended up with x equals 1 *10^-6. But what confuses me is why that is the amount of HCl you have to add, when the concentration of OH- ions is twice that (2x). I guess Im missing something here.😡

Hmmmmm...I guess because HCl only has 1 proton to donate and youve got 2OHs to neutralize? Not so sure😎

 

[_veo_]

thanks for the quick responses 😍
it does make more sense now..

 

[Manyak222]

Wait are we talking about the same problem (#1)?

Hmmmmm...I guess because HCl only has 1 proton to donate and youve got 2OHs to neutralize? Not so sure😎

 

[Manyak222]

Hook me up with an explanation coconut girl! 😉

 

[fancymylotus]

Hook me up with an explanation coconut girl! 😉

ok coconut boy.....i think you just needed to know the conc of the total OH ions and thats the amount of acid to add. quit thinking about it so much,sheesh.😎 😴 😀

 

[Manyak222]

I found x, that was easy. its 1 *10^-6. But kaplan says that the concentration of OH is twice that, 2x. Thus you'd need that many HCls (2*10^-6 moles.
Im sorry I guess either I dont get it or im right. (likely the former)

 

[fancymylotus]

I found x, that was easy. its 1 *10^-6. But kaplan says that the concentration of OH is twice that, 2x. Thus you'd need that many HCls (2*10^-6 moles.
Im sorry I guess either I dont get it or im right. (likely the former)

maybe...JUST MAYBE youre overthinking this all a lil bit 😉

 

[Manyak222]

well maybe once i get an explanation as to why im wrong ill forget about it😉

 

[fancymylotus]

well maybe once i get an explanation as to why im wrong ill forget about it😉

😴 😴 😴 😴 😴

 

[Manyak222]

im begining to think that im right.. In fact, i AM right. You need twice the amount of HCl due to OH concentration being 2 E -6 as kaplan says on page 285 in the middle.

 

[Anemone]

😴 😴 😴 😴 😴

No send me more!!