The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Ocr MCQ Help

50cm3 of 6.0moldm–3 HCl is mixed with 90cm3 of 3.0moldm–3 HNO3.What is the H+(aq) concentration in the resulting solution?

A 1.9moldm–3
B 2.1moldm–3
C 4.1moldm–3
D 4.5moldm–3

Answer = C - Can someone please explain why?

A 0.040moldm–3 solution of a weak monobasic acid is 1.0% dissociated.
What is the value of Ka for the acid?
A 2.0 × 10–7moldm–3
B 4.0 × 10–6moldm–3
C 4.0 × 10–4moldm–3
D 4.0 × 10–2moldm–3

Answer = B Why is the answer not C?

Thansk in advance for any help :smile:
(edited 5 years ago)
Reply 1
Avatar for jsg9
jsg9
14
For the first one:

You need to work out the total number of moles in the total volume of solution.
1) work out n of moles in each solution:
- 50/1000 x 6.0 = 0.3 mol HCl
- 90/1000 x 3.0 = 0.27 mol HNO3
2) add up your number of moles to work out the total number - 0.3 0.27 = 0.57 total moles of acid in solution
3) find the total volume of solution in dm^3
- 90 50/1000 = 0.14dm3
4) you have 0.57mol acid / H in 0.14dm^3 of solution so use your c = n/v equation to find the conc: 0.57 / 0.14 = 4.0714..... which rounds to 4.1mol dm^-3

You can assume the [H^ ] is the same as the conc of each acid as HCl and HNO3 are both strong acids.

For the second one:
Ka = conc of H squared / conc of HA
The acid is 1.0% dissociated, so 0.01 (1%) x 0.040 (the conc of the weak acid) = 0.0004

0.0004 = conc of H
So substitute into the equation for Ka

Ka = (0.0004)^2 / (0.04) = 4.0 x 10^-6 mol dm^-3

Hope that helps.

Edit: I know I should use square brackets in the Ka equation but tsr is omitting any text that I include within square brackets.
(edited 5 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you