1.What is the pH of a 0.1M of NaCl?
2. What volume of water must be added to 80ml of 12M of HCl to yield a solution the is 4.0M in HCl?
2. What volume of water must be added to 80ml of 12M of HCl to yield a solution the is 4.0M in HCl?
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It depends on the pH of the water used to make the solution of NaCl. Since Na+ is the cation of a strong base, and Cl- is the anion of a strong acid, neither hydrolyze (react with water) to alter the pH of the water.
Use the dilution equation:
mL(stock) x M(stock) = mL(dil) x M(dil)
solve for mL(dil)
mL(dil) = mL(stock) x M(stock) / M(dil)
mL(dil) = 80mL x 12M / 4.0M
mL(dil) = 240 mL
You might think that you would need to add 160 mL of water to the acid, but that may not be the case. You need to dilute the concentrated acid to 240 mL to make it 4.0M by adding however much water it takes. This is because when diluting a concentrated solution, the volumes are not additive.
The bottom line is that the question is flawed. A better question would be, "What will be the volume of the acid solution made by diluting 80.0 mL of 12M HCl with enough water to make it 4.0M?"
Use the dilution equation:
mL(stock) x M(stock) = mL(dil) x M(dil)
solve for mL(dil)
mL(dil) = mL(stock) x M(stock) / M(dil)
mL(dil) = 80mL x 12M / 4.0M
mL(dil) = 240 mL
You might think that you would need to add 160 mL of water to the acid, but that may not be the case. You need to dilute the concentrated acid to 240 mL to make it 4.0M by adding however much water it takes. This is because when diluting a concentrated solution, the volumes are not additive.
The bottom line is that the question is flawed. A better question would be, "What will be the volume of the acid solution made by diluting 80.0 mL of 12M HCl with enough water to make it 4.0M?"