Question

How many milliliters of 0.120M HCl are needed to completely neutralize 50.0 mL of 0.101 M Ba(OH)2 ?

Answer 1

Answer:

84mL

Explanation:

the neutralization reaction between HCl and Ba(OH)2 is

2HCl+Ba(OH)2 ----->BaCl2+2H2O

For HCl. For Ba(OH)2

M1=0.120M. M2=0.101M

V1=? V2=50mL

n1=2 moles. n2=1 mike

then as for the formula

$\frac{m1 \times v1}{n1} = \frac{m2 \times v2}{n2}$

0.120×V1/2 =50×0.101

V1 = 100×0.101/0.120

= 84.1mL

Answer 2

Milliliters of $0.120M HCl$ are needed to completely neutralize$50.0 mL$ of$0.101 M$ $Ba(OH)_2$  is  $84.2 mL.$

Step by step explanation:

• Equation of reaction :

                    $2HCl+Ba(OH)_2 = Bacl_2+2H_2O$

• First, calculate the number of moles of $Ba(OH)_2$   in $50.0 mL$of $0.101M$solution.

• $50.0\;mL\;\times\;(0.101\;mol\;/\;1000\;mL)\;=\;0.00505\;mol Ba(OH)_2$

• This tells us how many moles of $Ba(OH)_2$  must be neutralized.

• Next, we must calculate the number of moles of $HCl$ required to completely react with$0.00505$ moles of $Ba(OH)_2$.

• We will use the molar ratio derived from the balanced chemical equation($2 moles HCl.$to $1moles Ba(OH)_2$

• $0.00505 mol Ba(OH)_2 \times (2 mol HCl / 1 mol Ba(OH)_2 = 0.0101 mol HCl$

• This is the number of moles of $HCl$  required to neutralize the base $Ba(OH)_2$ Now, we need to figure out how many mL of the $0.120 M HCl$solution contain$0.0101 moles$ $HCl$ because this is the number of moles required to neutralize the base and the volume of solution that contains this number of moles.

• $0.0101 mol HCl\times(1000 mL / 0.120 mol HCl) = 84.2 mL of 0.120 M HCl$.

Therefore,$84.2 mL$ of the$0.120 M HCl$ solution are required to neutralized $50.0 mL$of the$0.101 M Ba(OH)_2$  solution.