Approximately how much water should be added to 10.0 mL of 12.0 M HCl so that it has the same pH as .90 M acetic acid(Ka= 1.8 x 10^-5)?
Calculate the (H^+) for 0.9 M acetic acid.
Then 10.0 mLof 12M x 12MHCl = ?mLH2O x Mfrom acetic acid.
Solve for mL H2O ahd subtract 10 mL from that to find how much to add.
Um , what grade chemistry is this for ?
since everything is a 1:1 ratio, you know that [acetic acid] = [H3O+]
forgot to add that..
wow i'm just forgetting every thing today:
when you find v2, don't forget to minus the initial volume!!! =) (since they are asking how much is added)
i got 29.80L
i want to find the density of acetic acid at 30 degree
To find out how much water should be added to 10.0 mL of 12.0 M HCl to match the pH of 0.90 M acetic acid, we need to consider the concept of pH and the dissociation of acids.
First, let's understand pH. pH is a measure of the acidity or alkalinity of a solution and is defined as the negative logarithm (base 10) of the concentration of hydrogen ions (H+). Mathematically, it is expressed as pH = -log[H+].
For acetic acid, the given pH is not directly provided. However, we are given the value of the acid dissociation constant (Ka = 1.8 x 10^-5). The dissociation equilibrium for acetic acid can be written as follows:
CH3COOH ⇌ CH3COO- + H+
The dissociation constant (Ka) is defined as the ratio of the concentrations of the products (CH3COO- and H+) to the concentration of the reactant (CH3COOH). In this case, we can consider the concentration of [H+] as the concentration of acetic acid [CH3COOH] since acetic acid is a weak acid and primarily exists as the undissociated molecule in solution.
Now, we need to find the concentration of HCl that would give us the same concentration of [H+] as that of acetic acid (0.90 M). To do this, we can set up an equation using the dissociation equilibrium expression of HCl:
HCl ⇌ H+ + Cl-
As HCl is a strong acid, it dissociates completely in solution, meaning that the concentration of H+ would be the same as the initial concentration of HCl ([H+] = [HCl]).
By comparing the two dissociation equilibria, we can equate the concentrations of H+ for acetic acid and HCl:
[H+] (acetic acid) = [H+] (HCl)
0.90 M = [H+] (HCl)
Since [H+] (HCl) = [HCl], the concentration of HCl we need to match the pH of acetic acid is also 0.90 M.
Now, let's calculate the volume of water needed to dilute 10.0 mL of 12.0 M HCl to obtain a final concentration of 0.90 M.
The formula for dilution is:
C1V1 = C2V2
Where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume.
Given:
C1 = 12.0 M
V1 = 10.0 mL
C2 = 0.90 M
V2 = ?
Rearranging the formula:
V2 = (C1V1) / C2
V2 = (12.0 M * 10.0 mL) / 0.90 M
V2 ≈ 133.33 mL
Therefore, approximately 133.33 mL of water should be added to 10.0 mL of 12.0 M HCl to obtain a solution with the same pH as 0.90 M acetic acid.
first find the [ace acid] (which is 0.0040~M)
use Ka constant formula to solve for [H3o+]
then use C1V1 = C2V2
where c1= [ace acid] = 0.004~M
c2= [initial HCl] = 12M
v1= (initial volume of HCl) = 0.01L
and you have to solve for v2
i think this is right lol
and roman, it's a Grade 12 + level question