can someone solve this chem problem???

[dsony2284]

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. If you add a drop (about 0.05 mL) of 1.0 M HCl to one liter of pure water (assume pH 7.0), the pH would be

 

[Danny289]

. If you add a drop (about 0.05 mL) of 1.0 M HCl to one liter of pure water (assume pH 7.0), the pH would be

Hi, approach for this problem :
M1V1=M(V1+V2)
V1= .05 is negible!
1=M(1000)====> M= 10^-3====> monoprotic Acid ==> N= M===> N=10^-3
then [H+]=10^-3==> [H+]=10^-PH ===>
PH=3
hopefuly it helped, good luck

 

[doc toothache]

. If you add a drop (about 0.05 mL) of 1.0 M HCl to one liter of pure water (assume pH 7.0), the pH would be

ml x M = ml2 x M2
0.05 x1 = 1000M2
M2= 5 x 10^5 (of HCl)
since HCl is 100% ionized [H+]=M2
pH= 4.3

 

[dsony2284]

thank u so much. yes the answer is 4.3 so great job guys...👍

 

[Danny289]

DOC is right .05 is neglible in right side of equation