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Chemistry 17 Online
OpenStudy (marmar10):

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.

OpenStudy (marmar10):

Na2O(aq)+H2O(l)→2NaOH(aq)

OpenStudy (marmar10):

@Hoslos

OpenStudy (marmar10):

@Cuanchi

imqwerty (imqwerty):

just find the molarity of [OH^{-} ] in NaOH and then take its negative log with base=10

OpenStudy (marmar10):

im still not getting the right answer... so i found M=.000161342. then i take the negative log of that? @imqwerty

imqwerty (imqwerty):

ok :) ima tell it step wise

imqwerty (imqwerty):

1st of all can u calculate the number of moles of NaOH that will be formed?

OpenStudy (marmar10):

yes i did that

imqwerty (imqwerty):

whats the value u got?

OpenStudy (marmar10):

.088738303

imqwerty (imqwerty):

yes correct :) can u tell this-> how many moles of (OH) are present in 0.0887 moles of NaOH?

OpenStudy (marmar10):

ohhh thats what i didnt do. there are 2

imqwerty (imqwerty):

there are 2?? what do u mean?

OpenStudy (marmar10):

wait nvm.. read that wrong.. wouldnt it just be the same?

imqwerty (imqwerty):

yes the moles will be the same :)

imqwerty (imqwerty):

next step is to calculate the molarity of OH

OpenStudy (marmar10):

i got .000161342 for that

imqwerty (imqwerty):

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 :)

OpenStudy (marmar10):

it says assume there is no volume change

OpenStudy (marmar10):

so wouldnt i keep it in mL?

imqwerty (imqwerty):

yes there is no volume change it means the volume is constant :) ok lemme give u an example :) can u write 1000grams as 1kilogram?

OpenStudy (marmar10):

1000 grams is 1 kilogram

imqwerty (imqwerty):

like this 1000ml=1Liter ^both are same so how many liters are equal to 550ml??

OpenStudy (marmar10):

.55 L

imqwerty (imqwerty):

yes :) we changed the volume from ml to L cuz we use litres to find out molarity so calculate the molarity now

OpenStudy (marmar10):

so now M= .16134236

imqwerty (imqwerty):

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

imqwerty (imqwerty):

here [OH] represents the concentration(molarity of OH)

OpenStudy (marmar10):

ok so i got pH= 13.208

imqwerty (imqwerty):

yea correct :)

OpenStudy (marmar10):

awesome!! thank you so much!! :)

imqwerty (imqwerty):

np B) yw (:

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