# Acids and bses

#### acetylmandarin

2+ Year Member
1 mL of a strong acid solution has a pH of 2.3. It is diluted with water to make a 100 mL solution. What is the new pH?

I don't really understand the concepts behind this question, could someone show how to work it out using definitions?

Also, with this question, and from my past few days of studying, I'm starting to get very worried. I am forgetting some of the most basic concepts from my prereq classes, and I have been studying for this exam for 3 months now. Nothing is really sticking; not now, and not back when I 'learned' these concepts during my classes......it seems like I'm running into a lot of trouble with every chapter

Last edited:
OP

#### acetylmandarin

2+ Year Member
I don't really understand the concepts behind this question, could someone show how to work it out using definitions?

I know that the original H3O+ concentration would be 10^-2.3, is there a way to solve that without a calculator? And then something to do with the solution now be 100x less concentrated?

Also, with this question, and from my past few days of studying, I'm starting to get very worried. I am forgetting some of the most basic concepts from my prereq classes, and I have been studying for this exam for 3 months now. Nothing is really sticking; not now, and not back when I 'learned' these concepts during my classes......it seems like I'm running into a lot of trouble with every chapter[/QUOTE]

#### aldol16

2+ Year Member
Okay, so this is a solute-solvent problem. The "solute" here is protons. You have a given number of protons and they are dissolved in a set amount of water. So you have 1 mL of a strong acid (so you can assume complete ionization) that has a pH of 2.3. Recall that pH is defined as -log[H+]. So [H+] is 10^-pH, or 10^-2.3 in this case. Remember that [H+] when in the context of pH is always expressed in units of molars. So you have 10^-2.3 M [H+]. Okay, now you have 1 mL of this solution, or 0.001 L. Remember the definition of molarity? Molarity is simply moles per liters. So you have 10^-2.3 moles/L * 0.001 L of protons, or around 5 x 10^-6 moles of protons.

Okay, you should know that an amount of something can't change if you just dilute it. So think of throwing a baseball into a beaker. You have one baseball in the beaker. Now take dump that beaker with the baseball in it and dump it into a swimming pool. You've just diluted the solution, but you still only have one baseball. The baseball is like the number of protons - it doesn't change by dilution. So you will always have 5 x 10^-6 moles of protons. Okay, so now you're taking those protons and putting them in 100 mL worth of solution. The new molarity is simply moles/liters, or 5 x 10^-6 moles/0.100 L, or 5 x 10^-5 M, or 50 micromolar.

An alternative route once you've calculated the initial concentration is to use M1*V1 = M2*V2.

OP

#### acetylmandarin

2+ Year Member
Okay, so this is a solute-solvent problem. The "solute" here is protons. You have a given number of protons and they are dissolved in a set amount of water. So you have 1 mL of a strong acid (so you can assume complete ionization) that has a pH of 2.3. Recall that pH is defined as -log[H+]. So [H+] is 10^-pH, or 10^-2.3 in this case. Remember that [H+] when in the context of pH is always expressed in units of molars. So you have 10^-2.3 M [H+]. Okay, now you have 1 mL of this solution, or 0.001 L. Remember the definition of molarity? Molarity is simply moles per liters. So you have 10^-2.3 moles/L * 0.001 L of protons, or around 5 x 10^-6 moles of protons.

Okay, you should know that an amount of something can't change if you just dilute it. So think of throwing a baseball into a beaker. You have one baseball in the beaker. Now take dump that beaker with the baseball in it and dump it into a swimming pool. You've just diluted the solution, but you still only have one baseball. The baseball is like the number of protons - it doesn't change by dilution. So you will always have 5 x 10^-6 moles of protons. Okay, so now you're taking those protons and putting them in 100 mL worth of solution. The new molarity is simply moles/liters, or 5 x 10^-6 moles/0.100 L, or 5 x 10^-5 M, or 50 micromolar.

An alternative route once you've calculated the initial concentration is to use M1*V1 = M2*V2.

Thank you. I know it's just a simple calculation, but I need more practice I guess...

OP

#### acetylmandarin

2+ Year Member
Okay, so this is a solute-solvent problem. The "solute" here is protons. You have a given number of protons and they are dissolved in a set amount of water. So you have 1 mL of a strong acid (so you can assume complete ionization) that has a pH of 2.3. Recall that pH is defined as -log[H+]. So [H+] is 10^-pH, or 10^-2.3 in this case. Remember that [H+] when in the context of pH is always expressed in units of molars. So you have 10^-2.3 M [H+]. Okay, now you have 1 mL of this solution, or 0.001 L. Remember the definition of molarity? Molarity is simply moles per liters. So you have 10^-2.3 moles/L * 0.001 L of protons, or around 5 x 10^-6 moles of protons.

Okay, you should know that an amount of something can't change if you just dilute it. So think of throwing a baseball into a beaker. You have one baseball in the beaker. Now take dump that beaker with the baseball in it and dump it into a swimming pool. You've just diluted the solution, but you still only have one baseball. The baseball is like the number of protons - it doesn't change by dilution. So you will always have 5 x 10^-6 moles of protons. Okay, so now you're taking those protons and putting them in 100 mL worth of solution. The new molarity is simply moles/liters, or 5 x 10^-6 moles/0.100 L, or 5 x 10^-5 M, or 50 micromolar.

An alternative route once you've calculated the initial concentration is to use M1*V1 = M2*V2.

Is there a quick way to write something like 10^-4.74 out in decimal form without a calculator?

#### aldol16

2+ Year Member
Is there a quick way to write something like 10^-4.74 out in decimal form without a calculator?
Yeah, your physics review book likely has a section on logs. So you know that 10^-4 is 0.0001 and 10^-5 is 0.00001. So you must conclude that 10^-4.74 is in between those values. For MCAT purposes, that's usually enough to get to the right answer. Now, if you want something closer to the correct value, recognize that -4.74 is closer to -5 than -4. So it's going to be very close to 0.00001 - the answer is 0.000018.

acetylmandarin