gen chem

[bxs023100]

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can someone help me in understanding how to do these:

1. when 50 g of HCl are dissolved in 100mL of H2O to make an acidic solution, what is the pOH of the resulting solution?

2. when 10 g of NaOH is dissolved in 100 mL of H2O to make a basic solution, what is the pH of the resulting solution?

 

[UNMPDS]

can someone help me in understanding how to do these:

1. when 50 g of HCl are dissolved in 100mL of H2O to make an acidic solution, what is the pOH of the resulting solution?

2. when 10 g of NaOH is dissolved in 100 mL of H2O to make a basic solution, what is the pH of the resulting solution?

Grams HCL to moles HCL to Molarity HCL (moles/liter (convert liter to milliliter)) to pH HCL or -log(HCL) to pOH HCL (pH - pOH= 14)

Do same for number 2

Make sense?

 

[bxs023100]

ya...thanks!!!

 

[scottyhoop]

Grams HCL to moles HCL to Molarity HCL (moles/liter (convert liter to milliliter)) to pH HCL or -log(HCL) to pOH HCL (pH - pOH= 14)

Do same for number 2

Make sense?

I'm good upto Molarity. Where after that? I get stuck on converting M to pH. Help an old man out that forgets acid/base chem!! 🙂
Scott

 

[littlemissy]

remember that only works for strong acids and bases

 

[Danny289]

Hi, guys
just I reviewed you Q & A

just pay attention to this Questions:
Add 50 gr Hcl to 100ml Water========> mean you are going to have solution more than 100ml ! ok how much is that? you must have valoum of 50 gr Hcl, Ok how we can have the volum of Hcl. you must have d( density for Hcl) d= m/v you will get Hcl volum
V(total)= vHcl + 100 ml
now we are in same page!!

and work for the rest of that I know you are better than me!
reminder: pay attention adding 50 gr Hcl or 10 gram NaoH to soultion, you are talking about molality not about Molarity
if you want to talk about molarity you must change the problom in this form😛ay attention to my changes in the original problom and you can do for second question too, with this change you don't need Hcl density. and you can solve problom easyly.
when 50 g of HCl are dissolved in H2O and make 100 ml acidic solution, what is the pOH of the resulting solution?

I am sorry for my long explanation. I am learning alot from you guys.
Good luck for your DAT

 

[Danny289]

to sccoty
just a little quote to my preivios answer,
I can say for this qustion POH must between 13 1nd 14 without solving the problom.
how?
assume; I have 3.66 gram HCl in 1000 ml soultion!!!( 0.1M)
I want to find PH & POH for this soultin:

We know: PH= -log[H+]
[H+]=0.1
PH=-log[0.1] ==> PH= -log10^-1 ====> -(-1)log10 and you know log10 is equal to 1 =========> PH=1

(I am sure you are good in "log" just i explaiened a little more to refresh our memories🙂 )

PH+POH=14
1+POH=14 =======>POH=13
Now imagine in our original problom acidity is much much more stronger!! result: we are going to have lower PH and Higher POH than our assumtion problom.