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A solution of acid is prepared for cleaning surgical instruments by adding 0.5L of 2mM HCL to 0.5L of pure water, which has a hydrogen ion concentration of 10-7 M. The initial pH of the pure water, then the pH after adding the HCL, are:
a) 7, then 3
b) 7, then 4
c) 7, then 1
d)14,then 3
e)14,then 4
Please answer this questions together with explanation. Thank you.
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pH= -log[H+]
The addition of same volume of the pure Water (pH=7) to the solution with pH= -log[0.002] initialy will be a new solution of pH= -log[0.002/2]= -log[0.001]= 3
So the new solution pH is 3.
So the answer is aaa
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I don't know how to calculate it.
Thanks for your explanation, michaelis.
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Michaelis can u please elaborate more in detail...ur explanation
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Evrerything is matter of this formula: pH= -log[H+]
Pure Water which is neutral has pH= -log[concentration in H+]= - log[0,0000001]= -[-7]=7
Solution of 2mMol has a concentration in H+ equal to 0.002 so pH= -log[0.002]
But went it got dilutated with same amount fo pure water its concentration fell down to half of intial which is 0.002/2= 0.001
So the pH of the new solution is -log[0.001] = 3
So the pure water of pH of 7 switch to pH= 3 when they added to it the solution.
NB log stands for decimal logarithm
Thanks
