A solution of hydrochloric acid of unknown concentration was titrated with 0.10M naoh.
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A solution of hydrochloric acid of unknown concentration was titrated with 0.10M naoh.

[From: ] [author: ] [Date: 12-07-09] [Hit: ]
Wats the pH of 2.00 M HC?pH = -log 2.00 = -0.pH values can be negative. They are not often used,......
a solution of hydrochloric acid of unknown concentration was titrated with 0.10M naoh. if a 100.ml sample of the hcl solution required exactly 10.ml of the naoh solution to reach the equivalence point, what was the ph of the hcl solution?

-
M1V1 = M2V2

(0.10 mol/L) (10 mL) = (x) (100 mL)

x = 0.010 M

pH = log [H+] = - log 0.01 = 2

"its obviously not negative one because ph isnt negative."

Wat's the pH of 2.00 M HC?

pH = -log 2.00 = -0.301

pH values can be negative. They are not often used, but they can exist.

-
Actually very concentrated strong acids can achieve a negative ph, consider the pH of 10M HCl:

-log(10) = -1

Anyway, back to the question:

You know how many moles of NaOH were titrated, and since the reaction:

H+ + OH- ---> HOH (H2O if you prefer)

has 1:1 stochiometry, you can use the conversion of # moles OH- titrated = # moles H+ consumed and vice versa.

.1M * .01L = .001 moles NaOH used.

.001 moles OH- = .001 moles H+, so:

.001/.1L = .01M

-log(.01) = 2

The answer is thus b.

-
0.10M of NaOH
Multiply the molarity of NaOH by the volume of NaOH
Molarity is measured in litres, volume has to be in litres too, hence I divided the mL by 1000 to get L (1L = 1000mL)
0.10M x 10 / 1000 = 0.001,mols of NaOH
Ratio 1:1
0.001mols of HCl
Divide the mols of HCl by the volume of HCl
0.001 / 100 / 1000 = 0.01M

Use the -log() formula
pH = -log(0.01)
pH = 2
Your answer is B
1
keywords: with,of,solution,acid,unknown,0.10,was,hydrochloric,titrated,naoh,concentration,A solution of hydrochloric acid of unknown concentration was titrated with 0.10M naoh.
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