The concentration of HCL is ____M?

A 25.0 mL sample of an HCL solution is titrated with a 0.139 M NaOH solution. The equivalence point is reached with 15.4 mL of base. The concentration of HCL is? Can someone please so me how to set this up... Thanks

Answer:
The reaction is:

NaOH + HCl >> NaCl + H20

0.139molesNaOH x 0.0154L=2.14x10-3 moles NaOH
....1 Liter

Considering you have a 1 to 1 ratio you have 2.14x10-3 moles of HCl

2.14x10-3moles/0.025L=8.56x10-... HCl
1ml = 1cm3

moles = (volume(cm3) x concentration)/1000

Moles of NaOH at equiv point = (15.4 x 0.139)/1000
= 2.1506 x 10^-3

Because at the equivalence point the moles of both solution are equal

2.1506 x 10^-3 = (25 x [HCl])/1000
2.1406 = 25[HCl]
[HCl] = 0.086moldm^-3
the simple answer is

(molarity 1)(volume 1)=(molarity 2)(volume 2)

so

(0.139 M NaOH)(0.0154L NaOH)=(x M HCl)(0.025L)

so

0.139*.0154/.025=0.085624 M HCl
0.139 m ---- 1000 ml
x----- 15.4

x=0.002085 -------15.4ml
.0834 M ---- 1000ml

Your HCL solution is 0.084 M

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