#1
question 3a(iii)

I just don't understand why they did that????????

Question paper
https://pmt.physicsandmathstutor.com...e%204%20QP.pdf

Marks scheme
https://pmt.physicsandmathstutor.com...e%204%20MS.pdf
0
2 years ago
#2
(Original post by Riannnne)
question 3a(iii)

I just don't understand why they did that????????

Question paper
https://pmt.physicsandmathstutor.com...e%204%20QP.pdf

Marks scheme
https://pmt.physicsandmathstutor.com...e%204%20MS.pdf
So 1.25x10^-3 mol of HxA reacts with 2.5x10^-3 mol of NaOH. You need to find how many moles of HxA react with NaOH. Divide the 2 values to find the molar ratio - 1 mol of HxA reacts with 2 mol of NaOH. This means in order for the acid to neutralise the alkali, the acid must supply 2 hydrogen ions (H + OH = H20) for each OH ion in the NaOH. Therefore x=2
0
2 years ago
#3
(Original post by Riannnne)
question 3a(iii)

I just don't understand why they did that????????

Question paper
https://pmt.physicsandmathstutor.com...e%204%20QP.pdf

Marks scheme
https://pmt.physicsandmathstutor.com...e%204%20MS.pdf
They said that x is the number of H+ ions that can be replaced by metal ions to form salts. The metal ions would be the sodium. In the calculation in the mark scheme, it shows that the moles of NaOH is two times greater than HxA so there two times the number of sodium ions compared to ions of A so two sodium ions must replace the two H+ ions to form the salt.
0
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