A 30.75 mL sample of 0.450 MKOH was titration with an unknown concentration of HCl. The endpoint was reached when 29.35 mL of NaOH were delivered. Find the concentration of the titrant
You've posted quite a lot of questions but they're all based on the same concept and formula.
These questions contain two compounds, a sample and a titrant.
Step 1: Write the balanced chemical equation for the reaction, and note down the stoichiometric coefficients of both the sample and the titrant.
Step 2:
Use the following formula:
(M1*V1)/(M2*V2) = (n1/n2)
Where,
M1 = Molarity of sample
V1 = Volume of sample
M2 = Molarity of titrant
V2 = Volume of titrant
n1 = stoichiometric coefficient of sample
n2 = stoichiometric coefficient of titrant
In each question posted, n1 and n2 must be obtained from the formula, three other values are given, and the remaining value must be found using the above formula.
Thank you, I’m a parent trying to help her son who’s struggling with this unit. The questions are from a worksheet that is homework. He’s asking me questions I can’t answer and I wanted to get the answers so I could check his work and redirect him if he went astray.
We usually don't work all the problems and let the "student/helper" check the values. Rather, if you plug the values into the formula provided by Aora, that will provide answers to all of these neutralization problems. If you have questions about this problem please don't hesitate to ask.
I showed the explanation to my son that you provided, Arora. He says that makes more sense and feels he can do it now. Eureka!! Let’s hope it sticks!!
I was told to tell you “thank you” from him.
Great, If you/he would like for us to check, please post your work (not just the answers). We shall be happy to check the numbers for you.
Realized I didn’t write the question correctly! My son pointed that out.
Corrected question: A 30.75 mL sample of 0.450 M KOH was titrated with an unknown concentration of HCl. The endpoint was reached when 22.60 mL of HCl were delivered. Find the concentration of the titrant.
Here’s his work:
HCl + KOH -> H2O + KCl
KOH = .45M (M1) 30.75 (V1)
HCl = (x)M (M2) 22.6 (V2)
(1) KCl = n1
(1) HCl = n2
(M1-V1)/(M2-V2) = (n1/n2)
(.45-30.75)/(x-22.6) = 1
13.8375 = 22.6x
.61 = x
HCl = .61M
0.61228 is what I have on my calculator which i would round to 0.612 M. Some profs are picky about significant figures and might count 0.61 wrong.. In multiplication and division the number of digits allowed in the answer is no more than the least in any of the values used in the calculation. If you typed in all of the values as they appear in the problem, there are 4 s.f. (significant figures) in 30.75, 3 s.f. in 0.450, and 4 s.f. in 22.60 so you are allowed 3 in the answer; thus 0.612 is what is reported. I might guess that the 0.450 had some other digit following since all of the other values have 4 s.f.
Ok, I’ll let him know! I’m just thrilled he was able to complete it! Thank you, thank you for your help!
To find the concentration of the titrant (HCl) from the given information, we need to set up a balanced chemical equation and use the concept of stoichiometry.
The balanced chemical equation for the reaction between KOH (potassium hydroxide) and HCl (hydrochloric acid) is:
KOH + HCl → KCl + H2O
From the equation, we can see that the ratio between KOH and HCl is 1:1. This means that for every mole of HCl, we need one mole of KOH to react completely.
First, let's calculate the number of moles of KOH used in the titration:
Moles of KOH = Volume of KOH (in L) × Concentration of KOH
Given:
Volume of KOH = 30.75 mL = 0.03075 L
Concentration of KOH = 0.450 M
Moles of KOH = 0.03075 L × 0.450 M = 0.0138375 moles (rounded to 4 decimal places)
Since the stoichiometry ratio of KOH to HCl is 1:1, the number of moles of HCl used in the titration is also 0.0138375 moles.
Now, let's find the concentration of HCl:
Concentration of HCl = Moles of HCl / Volume of HCl (in L)
Given:
Volume of HCl = 29.35 mL = 0.02935 L
Concentration of HCl = 0.0138375 moles / 0.02935 L = 0.471 M (rounded to 3 decimal places)
Therefore, the concentration of the titrant (HCl) is approximately 0.471 M.