What is the pH of the solution prepared by allowing 2.75 g of Na2O to react with 550.0 mL of water? Assume that there is no volume change.
Na2O(aq)+H2O(l)→2NaOH(aq)
@Hoslos
@Cuanchi
just find the molarity of [OH^{-} ] in NaOH and then take its negative log with base=10
im still not getting the right answer... so i found M=.000161342. then i take the negative log of that? @imqwerty
ok :) ima tell it step wise
1st of all can u calculate the number of moles of NaOH that will be formed?
yes i did that
whats the value u got?
.088738303
yes correct :) can u tell this-> how many moles of (OH) are present in 0.0887 moles of NaOH?
ohhh thats what i didnt do. there are 2
there are 2?? what do u mean?
wait nvm.. read that wrong.. wouldnt it just be the same?
yes the moles will be the same :)
next step is to calculate the molarity of OH
i got .000161342 for that
yeah this is where u made the mistake when we calculate molarity we take the volume in litres but what we have is 550ml convert it into litres and then calculate :)
it says assume there is no volume change
so wouldnt i keep it in mL?
yes there is no volume change it means the volume is constant :) ok lemme give u an example :) can u write 1000grams as 1kilogram?
1000 grams is 1 kilogram
like this 1000ml=1Liter ^both are same so how many liters are equal to 550ml??
.55 L
yes :) we changed the volume from ml to L cuz we use litres to find out molarity so calculate the molarity now
so now M= .16134236
yes :) now u have the molarity we can calculate pOH \[p^{OH}=-\log[OH]\] after calculating pOH we use this equation to get our pH-\[p^{H}+p^{OH}=14\] substitute pOH that u get in this eq to get pH
here [OH] represents the concentration(molarity of OH)
ok so i got pH= 13.208
yea correct :)
awesome!! thank you so much!! :)
np B) yw (:
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