## 6B.3

Kelly Cai 4D
Posts: 55
Joined: Sat Jul 20, 2019 12:17 am

### 6B.3

A careless laboratory technician wants to prepare 200.0 mL of a 0.025 m HCl(aq) solution but uses a volumetric flask of volume 250.0 mL by mistake. (a) What would the pH of the desired solution have been? (b) What will be the actual pH of the solution as prepared?

DesireBrown1J
Posts: 98
Joined: Wed Sep 18, 2019 12:18 am

### Re: 6B.3

Can someone explain to me what I did wrong for this problem?

a) M=0.025 mol HCl/0.200 L =0.125 mol*L-1
pH=-log(0.125)=0.903?
But the solutions manual says 1.6

Connor Ho 1B
Posts: 102
Joined: Sat Aug 17, 2019 12:17 am

### Re: 6B.3

In this question, we are asked for two things:

1) the desired pH
2) the actual pH

For 1)
pH is the concentration of hydrogen atoms in solution. Because the equation for HCL is the following:

HCl --> H+ + Cl-

We can assume that since the concentration of HCl 0.025M, the concentration of H+ is also 0.025 M.

From here, we just do the equation for pH: -log(0.025) = 1.6 pH.

For 2)
Since we know that 200mL of HCl would give us 0.025M HCl, we can use the dilution equation to find the final molarity when using a 250mL volumetric flask:

M1V1 = M2V2

(0.025M)(.200L) = (M2)(.250L)
(0.025M)(.200L)/(.250L) = M2
M2 = 0.02 M

You would then take this answer and plug it into the pH equation: -log(0.02) = 1.7 pH.

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