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If 6 milliliters of 1M HCl is exactly neutralized by 3 milliliters of KOH, the molarity of the KOH is

a) 1M
b)3M
c)2M
d)9M

1 Answer

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Step-by-step explanation:

KOH molarity calculation.

Durga thapaliya Thapaliya

If 6 milliliters of 1M HCl is exactly neutralized by 3 milliliters of KOH, the molarity of the KOH is

a) 1M

b)3M

c)2M

d)9M

The balanced chemical equation for the reaction between HCl and KOH is:

HCl + KOH → KCl + H2O

From the equation, we can see that the mole ratio between HCl and KOH is 1:1. Therefore, the number of moles of HCl in 6 mL of 1M HCl is:

moles of HCl = volume of HCl × concentration of HCl

= 6 mL × 1 mol/L

= 6 × 10^-3 mol

Since the number of moles of HCl is equal to the number of moles of KOH, we can use the following formula to calculate the molarity of KOH:

molarity of KOH = moles of KOH / volume of KOH

We are given that 3 mL of KOH is used to neutralize the HCl. Therefore, the number of moles of KOH is:

moles of KOH = moles of HCl = 6 × 10^-3 mol

Substituting the values in the above formula, we get:

molarity of KOH = moles of KOH / volume of KOH

= 6 × 10^-3 mol / 3 mL

= 2M

Therefore, the molarity of KOH is 2M, and the correct option is (c).

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