Calculate the pH and pOH of each of the following aqueous solutions of strong acid or base:
(d) 2.00 mL of 0.175 m KOH(aq) after dilution to 0.500 L
I understand the concept of finding pH and pOH but I am stuck on this question because I do understand how they got the molarity for the dilution. In the solutions manual they did (2.00ml/500ml)*.175mol/L to get the new molarity. I am confused about how this follows the rules of molarity.
12.29d
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 23858
- Joined: Thu Aug 04, 2011 1:53 pm
- Has upvoted: 1253 times
Re: 12.29d
This basically used M1V1 = M2V2. New concentration is M2 = M1V1/V2. V1/V2 represents your 2.00/500. M1 would be 0.175.
-
- Posts: 19
- Joined: Fri Sep 29, 2017 7:07 am
Re: 12.29d
You need to use the formula M1V1=M2V2 to get the new concentration of KOH, which also equals to the concentration of OH- because KOH is a strong base.
Re: 12.29d
SO it would be (0.002)(.175)=(0.0005)(x)
X=0.7
-log(0.7)=0.15 but the answer is 3.15
what am I doing wrong?
X=0.7
-log(0.7)=0.15 but the answer is 3.15
what am I doing wrong?
-
- Posts: 105
- Joined: Fri Sep 24, 2021 7:18 am
Re: 12.29d
Katie 3H wrote:SO it would be (0.002)(.175)=(0.0005)(x)
X=0.7
-log(0.7)=0.15 but the answer is 3.15
what am I doing wrong?
Hi, I hope this isn't too late but I believe that for:
(0.002)(.175)=(0.0005)(x)
it should be
(0.002)(.175)=(0.5)(x)
0.5 is already in liters :)
Return to “Calculating pH or pOH for Strong & Weak Acids & Bases”
Who is online
Users browsing this forum: No registered users and 7 guests