Need help with a chemistry question (1 Viewer)

[Tecumseh]

Paging Mr. Pickles...

If the following solutions are mixed, would the resulting solution be acidic, basic, or neutral?
a. 50.0 mL of 0.100 hydrobromic acid and 30.0 mL of 0.200 M potassium hydroxide

Seems like it should be a simple matter of (.1)(.05) < (.2)(.030), which would make it a base, but my answer key says that there are more H+ ions than OH- ions, so it's acidic. Why?

 

[meatwad1]

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount, so it is an acid.

 

[Tecumseh]

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount, so it is an acid.

Can you break that down? Am I supposed to balance the equation somehow?

ETA: And where are you getting sodium acetate? Wouldn't the salt be potassium bromide?

 

[Mr. Pickles]

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount, so it is an acid.

Can you break that down? Am I supposed to balance the equation somehow?

ETA: And where are you getting sodium acetate? Wouldn't the salt be potassium bromide?

I think he's messing with you.

As you're likely aware, HBr is a strong acid and KOH a strong base. No need for Ka nonsense here. Full dissociation.

50 mL of a 0.1 M HBr solution yields 5 mmol of H+ ions. 30 mL of a 0.2 M KOH solution yields 6 mmol of OH- ions. There is an excess of 6 mmol - 5 mmol = 1 mmol OH- ions, thus the solution is basic with a pH of 14 - (-log(0.001 mol OH-)/(0.08 L)) = 12.1.

 

[Officer Pete Malloy]

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount, so it is an acid.

Can you break that down? Am I supposed to balance the equation somehow?

ETA: And where are you getting sodium acetate? Wouldn't the salt be potassium bromide?

I think he's messing with you.

As you're likely aware, HBr is a strong acid and KOH a strong base. No need for Ka nonsense here. Full dissociation.

50 mL of a 0.1 M HBr solution yields 5 mmol of H+ ions. 30 mL of a 0.2 M KOH solution yields 6 mmol of OH- ions. There is an excess of 6 mmol - 5 mmol = 1 mmol OH- ions, thus the solution is basic with a pH of 14 - (-log(0.001 mol OH-)/(0.08 L)) = 12.1.

Just came here to post this.

 

[kutta]

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount, so it is an acid.

Can you break that down? Am I supposed to balance the equation somehow?ETA: And where are you getting sodium acetate? Wouldn't the salt be potassium bromide?

I think he's messing with you.

As you're likely aware, HBr is a strong acid and KOH a strong base. No need for Ka nonsense here. Full dissociation.

50 mL of a 0.1 M HBr solution yields 5 mmol of H+ ions. 30 mL of a 0.2 M KOH solution yields 6 mmol of OH- ions. There is an excess of 6 mmol - 5 mmol = 1 mmol OH- ions, thus the solution is basic with a pH of 14 - (-log(0.001 mol OH-)/(0.08 L)) = 12.1.

Just came here to post this.

Yeah, me too! Weird...

 

[Tecumseh]

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount, so it is an acid.

Can you break that down? Am I supposed to balance the equation somehow?

ETA: And where are you getting sodium acetate? Wouldn't the salt be potassium bromide?

I think he's messing with you.

As you're likely aware, HBr is a strong acid and KOH a strong base. No need for Ka nonsense here. Full dissociation.

50 mL of a 0.1 M HBr solution yields 5 mmol of H+ ions. 30 mL of a 0.2 M KOH solution yields 6 mmol of OH- ions. There is an excess of 6 mmol - 5 mmol = 1 mmol OH- ions, thus the solution is basic with a pH of 14 - (-log(0.001 mol OH-)/(0.08 L)) = 12.1.

Yeah, I worked through it. I had misread the answer key.

Meatwad, DIAF.

 

[Jules Winnfield]

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount, so it is an acid.

Can you break that down? Am I supposed to balance the equation somehow?ETA: And where are you getting sodium acetate? Wouldn't the salt be potassium bromide?

I think he's messing with you.

As you're likely aware, HBr is a strong acid and KOH a strong base. No need for Ka nonsense here. Full dissociation.

50 mL of a 0.1 M HBr solution yields 5 mmol of H+ ions. 30 mL of a 0.2 M KOH solution yields 6 mmol of OH- ions. There is an excess of 6 mmol - 5 mmol = 1 mmol OH- ions, thus the solution is basic with a pH of 14 - (-log(0.001 mol OH-)/(0.08 L)) = 12.1.

Beat me to it

 

[meatwad1]

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount, so it is an acid.

Can you break that down? Am I supposed to balance the equation somehow?

ETA: And where are you getting sodium acetate? Wouldn't the salt be potassium bromide?

I think he's messing with you.

As you're likely aware, HBr is a strong acid and KOH a strong base. No need for Ka nonsense here. Full dissociation.

50 mL of a 0.1 M HBr solution yields 5 mmol of H+ ions. 30 mL of a 0.2 M KOH solution yields 6 mmol of OH- ions. There is an excess of 6 mmol - 5 mmol = 1 mmol OH- ions, thus the solution is basic with a pH of 14 - (-log(0.001 mol OH-)/(0.08 L)) = 12.1.

Yeah, I worked through it. I had misread the answer key.

Meatwad, DIAF.

Die in a Fire? Really?

 

[Raider Nation]

Pickles should cook meth in an RV.

 

[Tecumseh]

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount, so it is an acid.

Can you break that down? Am I supposed to balance the equation somehow?

ETA: And where are you getting sodium acetate? Wouldn't the salt be potassium bromide?

I think he's messing with you.

As you're likely aware, HBr is a strong acid and KOH a strong base. No need for Ka nonsense here. Full dissociation.

50 mL of a 0.1 M HBr solution yields 5 mmol of H+ ions. 30 mL of a 0.2 M KOH solution yields 6 mmol of OH- ions. There is an excess of 6 mmol - 5 mmol = 1 mmol OH- ions, thus the solution is basic with a pH of 14 - (-log(0.001 mol OH-)/(0.08 L)) = 12.1.

Yeah, I worked through it. I had misread the answer key.

Meatwad, DIAF.

Die in a Fire? Really?

Yes, really.

 

[Dan Lambskin]

:subscribe:

 

[Guest]

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount, so it is an acid.

Can you break that down? Am I supposed to balance the equation somehow?ETA: And where are you getting sodium acetate? Wouldn't the salt be potassium bromide?

I think he's messing with you.

As you're likely aware, HBr is a strong acid and KOH a strong base. No need for Ka nonsense here. Full dissociation.

50 mL of a 0.1 M HBr solution yields 5 mmol of H+ ions. 30 mL of a 0.2 M KOH solution yields 6 mmol of OH- ions. There is an excess of 6 mmol - 5 mmol = 1 mmol OH- ions, thus the solution is basic with a pH of 14 - (-log(0.001 mol OH-)/(0.08 L)) = 12.1

Yeah, I worked through it. I had misread the answer key.Meatwad, DIAF.

Die in a Fire? Really?

Yes, really.

Wouldn't he suffer more if he just GRBBIAF?

 

[Parmcat]

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount, so it is an acid.

Can you break that down? Am I supposed to balance the equation somehow?

ETA: And where are you getting sodium acetate? Wouldn't the salt be potassium bromide?

I think he's messing with you.

As you're likely aware, HBr is a strong acid and KOH a strong base. No need for Ka nonsense here. Full dissociation.

50 mL of a 0.1 M HBr solution yields 5 mmol of H+ ions. 30 mL of a 0.2 M KOH solution yields 6 mmol of OH- ions. There is an excess of 6 mmol - 5 mmol = 1 mmol OH- ions, thus the solution is basic with a pH of 14 - (-log(0.001 mol OH-)/(0.08 L)) = 12.1.

This guy knows ####!

Wicked Smaht

 

[Tom Hagen]

Let's talk chemical engineering!!!!!

 

[eoMMan]

Let's talk chemical engineering!!!!!

I could cut diamonds over here.