A 30.0 cm3 sample of a 0.480 mol dm–3 solution of potassium hydroxide was partially neutralised by the addition of 18.0 cm3 of a 0.350 mol dm–3 solution of sulphuric acid. (i) Calculate the initial number of moles of potassium hydroxide. ........................................................................................................................... ........................................................................................................................... (ii) Calculate the number of moles of sulphuric acid added. ........................................................................................................................... ........................................................................................................................... (iii) Calculate the number of moles of potassium hydroxide remaining in excess in the solution formed. ........................................................................................................................... ........................................................................................................................... (iv) Calculate the concentration of hydroxide ions in the solution formed. ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... (v) Hence calculate the pH of the solution formed. Give your answer to two decimal places.
How would you do the final part? The MS uses the value of kw as 10^-14 however the question doesnt state its at room temp. Initially i just divided the concentration of OH- by 2 as its a dirpotic acid however this was wrong
I worked out the moles of both and then did the ice box. x2 the moles of h2s04 because its a 1;2 ratio and then got excess moles of koh which was 1.8 something. convert that to concentration and then use kw which i just assumed was 1x10-14.
I worked out the moles of both and then did the ice box. x2 the moles of h2s04 because its a 1;2 ratio and then got excess moles of koh which was 1.8 something. convert that to concentration and then use kw which i just assumed was 1x10-14.
Is it okay to assume as i checked if their was any mention of room temp or 298k but there wasnt
i think youre doing an old paper? correct me if im wrong, but they will most likely tell you.
Yes its an old paper and there was no mention of it so i tried the other way, diprotic acid so 2 moles of OH- is one mole of H+ so i just divided the moles of OH- by 2 and subbed it into the PH formula, do you know why this is wrong?
For my exam board (AQA) you are supposed to know what kw is at 298k. If they do not give you a temperature, I think you assume that it was done under standard conditions (which is 298k). Unless you are supposed to know multiple kw values, they would have to tell you the different temperature and give you a new kw value for that temperature.
For my exam board (AQA) you are supposed to know what kw is at 298k. If they do not give you a temperature, I think you assume that it was done under standard conditions (which is 298k). Unless you are supposed to know multiple kw values, they would have to tell you the different temperature and give you a new kw value for that temperature.
for the final part how did you know kw was = to 10^-14
That's something we have to learn at Edexcel that kw = 1×10-14
There's another method you could use as well though, you can -log[OH-] which is -log[0.0375] = 1.4259687 then minus this from 14 (max pH) So 14 - 1.4259687= 12.57 which is the same answer