Yash12345
Badges: 9
Rep:
?
#1
Report Thread starter 3 weeks ago
#1
24.2cm^3 of 0.2M HCl was neutralised by 25cm^3 of an alkaline solution. The alkaline solution contained a 16.47g mixture of sodium and rubidium hydroxide dissolved to give 1000cm^3 of solution. Calculate the mass of NaOh in the mixture.

I have caluclated the moles of HCl (0.0242 x 0.2) but I don't know what to do next.
Any help appreciated.
0
reply
charco
Badges: 17
Rep:
?
#2
Report 3 weeks ago
#2
(Original post by Yash12345)
24.2cm^3 of 0.2M HCl was neutralised by 25cm^3 of an alkaline solution. The alkaline solution contained a 16.47g mixture of sodium and rubidium hydroxide dissolved to give 1000cm^3 of solution. Calculate the mass of NaOh in the mixture.

I have caluclated the moles of HCl (0.0242 x 0.2) but I don't know what to do next.
Any help appreciated.
Cracking question Grommit.

So you have the moles of acid = 4.84 x 10-3

And you know that both of the bases are strong and provide hydroxide ions.

The total moles of hydroxide ions must equal the acid ions = 4.84 x 10-3

If you let the moles of sodium hydroxide = x, then the moles of rubidium hydroxide = total moles of hydroxide - x

You also know the mass of the mixture in 25 ml = 16.47/40 = 0.4118g

if the moles of NaOH = x then the mass due to NaOH = 40x and the mass due to RbOH = (total moles - x)(85.56 + 17)

Hence:

0.4118 = 40x + (total moles - x)(85.56 + 17)

can you take it from here?
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (62)
15.12%
I'm not sure (14)
3.41%
No, I'm going to stick it out for now (139)
33.9%
I have already dropped out (6)
1.46%
I'm not a current university student (189)
46.1%

Watched Threads

View All