Calculate the pH of each of the following strong acid solutions:
a) 1.02 g of HNO3 in 250 mL of solution
b) 2.00mL of 0.500 M HClO4 diluted in 50.0 mL
c) a solution formed by mixing 10.0 mL of 0.0100M HBr with 20.0 mL of 2.50x10^-3 M HCl
a) 1.02 g of HNO3 in 250 mL of solution
b) 2.00mL of 0.500 M HClO4 diluted in 50.0 mL
c) a solution formed by mixing 10.0 mL of 0.0100M HBr with 20.0 mL of 2.50x10^-3 M HCl
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This is about to be intense! Ready? Here we go...
A.) First you have to recognize that HNO3 is a strong acid so if you just take the negative log of it's concentration you can find pH. Well how do you find its concentration? You just find its molarity (mol/L)!
1.02gHNO3 x (1molHNO3/63gHNO3) = 0.0162 mol HNO3 <----mol, now just need L
To find Liters, convert 250 mL of soln...should get 0.250 L
Molarity = mol/L = .0162molHNO3 / 0.250L = .0648 M <---concentration of HNO3...now just take -log
-log(.0648)= 1.19 <---- pH of HNO3 (makes sense for pH to be low because HNO3 is a strong acid)
B.) HClO4 is again another strong acid. So to find molarity you need what? Thats right, concentration of HClO4! Here we go...
This time to get moles you'll need to realize that the 0.500 M HClO4 is actually (.500mol/L) so to get mol youll need to multiply by liters. So 2 mL= .002L x .500 M = .001 mol HClO4!
Now this is diluted (means you added in water, or liters of water). So youll need to get those moles and divide by the liters of how much water you diluted it with to find concentration. So you have (.001mol HClO4/ .05L) = concentration of HClO4 = .02 M HClO4
Now to find pH you just take -log (.02) because HClO4 is a strong acid. You should get pH = 1.70. Makes sense for the pH to be low because HClO4 is a strong acid.
C. Now this ones a bit more tricky. Here you will have to find moles of both HBr and HCl first then add them. Then divide the total number of moles by the total volume (10mL+20mL=30mL=.03L). Im tired so ill let you do this yourself :]
A.) First you have to recognize that HNO3 is a strong acid so if you just take the negative log of it's concentration you can find pH. Well how do you find its concentration? You just find its molarity (mol/L)!
1.02gHNO3 x (1molHNO3/63gHNO3) = 0.0162 mol HNO3 <----mol, now just need L
To find Liters, convert 250 mL of soln...should get 0.250 L
Molarity = mol/L = .0162molHNO3 / 0.250L = .0648 M <---concentration of HNO3...now just take -log
-log(.0648)= 1.19 <---- pH of HNO3 (makes sense for pH to be low because HNO3 is a strong acid)
B.) HClO4 is again another strong acid. So to find molarity you need what? Thats right, concentration of HClO4! Here we go...
This time to get moles you'll need to realize that the 0.500 M HClO4 is actually (.500mol/L) so to get mol youll need to multiply by liters. So 2 mL= .002L x .500 M = .001 mol HClO4!
Now this is diluted (means you added in water, or liters of water). So youll need to get those moles and divide by the liters of how much water you diluted it with to find concentration. So you have (.001mol HClO4/ .05L) = concentration of HClO4 = .02 M HClO4
Now to find pH you just take -log (.02) because HClO4 is a strong acid. You should get pH = 1.70. Makes sense for the pH to be low because HClO4 is a strong acid.
C. Now this ones a bit more tricky. Here you will have to find moles of both HBr and HCl first then add them. Then divide the total number of moles by the total volume (10mL+20mL=30mL=.03L). Im tired so ill let you do this yourself :]