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pH Calculation (2 Viewers)

micuzzo

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i need help with this one... caan someone please explain how to do it.

20mL of 0.08mol /L HCl is mixed with 30mL of 0.05mol/L NaOH. What is resultant pH??
 

lolokay

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a neutralisation reaction will occur between the NaOH and HCl, so just look at how much of each you have, which is in excess, and what the pH will be due to this excess after neutralisation has completed (I think)
 

gunne

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Do you mean find molarity or find pH... I think you must mean molarty but i might be wrong
 

micuzzo

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well u need to find molarity to find the ph... can someone do it
 

GUSSSSSSSSSSSSS

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sorry i cbb doin it
but wat u gotta do:

write the reaction between HCl and NaOH

find the limiting reagent

then find the moles of the substance in excess, that is the moles of the limiting reagent subtracted from the moles of the substance in excess

then calculate the concentration of that substance in solution, so dividing it by the volume of the two solutions added together

then u find the concentration of H+ ions/OH- ions, which in this case will be equal to the concentration u just found

and then sub in the formula: -logH+

XDDDD
 

gunne

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wait you already have molarity of both...
pH = -log(10) (Hydrogen ion concentration)...
does that help
 

undalay

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20mL of 0.08mol /L HCl is mixed with 30mL of 0.05mol/L NaOH. What is resultant pH??

nHCl = VC = 0.02 x 0.08 = 0.0016 moles
nNaOH = VC = 0.03 x 0.05 = 0.0015

H+ + OH- => H2O (Hydrogen ions and hydroxide ions annihilate.

0.0016-0.0015

remaining = 0.0001 moles of H+
total volume = 0.02+0.03 = 0.05

concentration of H+ remaining = n/v = 0.0001/0.05

-log [H+] = pH

pH = 2.7
 

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