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Marta_Voda [28]
3 years ago
15

A student neutralized 16.4 milliliters of HCl by adding 12.7 milliliters of 0.620 M KOH. What was the molarity of the HCl acid?

Chemistry
1 answer:
Katen [24]3 years ago
4 0
V ( HCl ) = 16.4 mL / 1000 => 0.0164 L

M( HCl) = ?

V( KOH) = 12.7 mL / 1000 => 0.0127 L

M(KOH) = 0.620 M

Number of moles KOH:

n = M x V

n = 0.620 x 0.0127

n = 0.007874 moles of KOH

number of moles HCl :

<span>HCl + KOH = H2O + KCl
</span>
1 mole HCl ------ 1 mole KOH
<span>? mole HCl--------0.007874 moles KOH
</span>
moles HCl = 0.007874 * 1 / 1

= 0.007874 moles of HCl

M = n / V

M = 0.007874 / <span>0.0164

</span>= 0.480 M

Answer (2)

hope this helps!

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