i am stuck on this irritating problem.. all because of astupid number.. i will w

Question

i am stuck on this irritating problem.. all because of astupid number.. i will write the question here and the solution too.. What is the pH of 0.10 mol dm-3 ethanoic acid ,gien that 1.3 percent of the ethanoic acid molecules havedissociated? Answer: Given that only 1.3 per cent of ethanoic acidmoleucles dissociate, then the concentration of ethanoic acid thathas dissociated must be 1.3 per cent of 0.100 mol dm-3 ,which is 1.3 * 10-3mol dm -3 Substituting this value into the equation for pH given: pH = -log10(1.3*10-3)= 2.9 SO MY PROBLEM IS: wheeeere does this 10-3 comefrom?? because 0.100 is not equal to 10-3 ..10-3 is equal to 0.001.. so how did they come upwith this value , 10-3 ???? IS IT just some mistake my textbook have done?! they did thesame kind of problem in the same way the next page.. and i stilldont get how they came up with this value!! PLEEEASE PLEASE HELP ME NOW! otherwise, ill be stuckwith this all night! must SOLVE IT! i am stuck on this irritating problem.. all because of astupid number.. i will write the question here and the solution too.. What is the pH of 0.10 mol dm-3 ethanoic acid ,gien that 1.3 percent of the ethanoic acid molecules havedissociated? Answer: Given that only 1.3 per cent of ethanoic acidmoleucles dissociate, then the concentration of ethanoic acid thathas dissociated must be 1.3 per cent of 0.100 mol dm-3 ,which is 1.3 * 10-3mol dm -3 Substituting this value into the equation for pH given: pH = -log10(1.3*10-3)= 2.9 SO MY PROBLEM IS: wheeeere does this 10-3 comefrom?? because 0.100 is not equal to 10-3 ..10-3 is equal to 0.001.. so how did they come upwith this value , 10-3 ???? IS IT just some mistake my textbook have done?! they did thesame kind of problem in the same way the next page.. and i stilldont get how they came up with this value!! PLEEEASE PLEASE HELP ME NOW! otherwise, ill be stuckwith this all night! must SOLVE IT! Substituting this value into the equation for pH given: pH = -log10(1.3*10-3)= 2.9 SO MY PROBLEM IS: wheeeere does this 10-3 comefrom?? because 0.100 is not equal to 10-3 ..10-3 is equal to 0.001.. so how did they come upwith this value , 10-3 ???? IS IT just some mistake my textbook have done?! they did thesame kind of problem in the same way the next page.. and i stilldont get how they came up with this value!! PLEEEASE PLEASE HELP ME NOW! otherwise, ill be stuckwith this all night! must SOLVE IT!

Explanation / Answer

1.3 per cent of 0.100 mol dm-3 , which is 1.3* 10-3mol dm -3 SO MY PROBLEM IS: wheeeere does this 10-3 comefrom?? because 0.100 is not equal to 10-3 ..10-3 is equal to 0.001.. so how did they come upwith this value , 10-3 ????


the textbook did the problem correctly.
must be 1.3 per cent of 0.100 mol dm-3 , which is 1.3 *10-3mol dm -3

1.3% of 0.100 mol dm-3 is actually 1.3 *10-3mol dm -3

solution:
1.3% = 0.013

hence 1.3% of 0.100 mol dm-3 = 0.013 x 0.100 moldm-3 = 0.0013 moldm-3 = 1.3 * 10-3mol dm -3 SO MY PROBLEM IS: wheeeere does this 10-3 comefrom?? because 0.100 is not equal to 10-3 ..10-3 is equal to 0.001.. so how did they come upwith this value , 10-3 ????


the textbook did the problem correctly.
must be 1.3 per cent of 0.100 mol dm-3 , which is 1.3 *10-3mol dm -3

1.3% of 0.100 mol dm-3 is actually 1.3 *10-3mol dm -3

solution:
1.3% = 0.013

hence 1.3% of 0.100 mol dm-3 = 0.013 x 0.100 moldm-3 = 0.0013 moldm-3 = 1.3 * 10-3mol dm -3

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