Respuesta :
The molarity of the lake water is 0.00001 M and the pH of lake water is 5.
The lake water is acidic.
Explanation:
Data given:
molarity of base solution Mbase = 0.1 M
volume of the base solution Vbase = 0.1 ml or 0.0001 litre
volume of lake water Vlake = 1000ml or 1 litre
molarity of the lake water, Mlake = ?
Using the formula for titration:
Mbase X Vbase = Mlake X
Mlake = [tex]\frac{Mbase X Vbase }{Vlake}[/tex]
Putting the values in the equation:
Mlake = [tex]\frac{0.0001 X 0.1}{1}[/tex]
Mlake = 0.00001 M
The pH of the lake water will be calculated by using the following formula:
pH = - [tex]log_{10}[/tex] [[tex]H^{+}[/tex]]
pH = -[tex]log_{10}[/tex] [ 0.00001]
pH = 5
a. Molarity =0.00001 M
b. pH of lake water =5
Given:
Molarity of base solution [tex]M_{base}[/tex] = 0.1 M
Volume of the base solution [tex]V_{base}[/tex] = 0.1 ml or 0.0001 litre
Volume of lake water [tex]V_{lake}[/tex] = 1000ml or 1 litre
Molarity of the lake water, [tex]M_{lake}[/tex] = ?
a. Using the formula for titration:
M₁V₁=M₂V₂
So,
[tex]M_{base}* V_{base}= M_{lake}*V_{lake}\\\\M_{lake}=\frac{M_{base}* V_{base}}{V_{lake}} \\\\M_{lake}=\frac{0.0001*0.1}{1} \\\\M_{lake}=0.00001M[/tex]
The molarity of lake is 0.00001M.
b. The pH of the lake water will be calculated by using the following formula:
[tex]pH=-log[H^+]\\\\pH=-log[0.00001]\\\\pH=5[/tex]
The pH of the lake water is 5.
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