Question

A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resulting solution? (Assume that the volumes are additive.) 0.10 M A 0.19 M B 0.28 M C 0.40 M D 0.55 M

Answer 1

Answer:

0.19 M is the correct answers

Explanation:

by

N1V1 + N2V2 = N3(V1+ V2)

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Answer 2

The molar concentration of OH-(aq) in the resulting solution is 0.28 M .

Given :

Volume of 0.25 M KOH is 40 ml .

Volume of 0.15 M $Ba(OH)_2$ is 60 ml .

Number of milli moles of $OH^-$ ions in KOH , $n_1=0.25\times 40=10$

Number of milli moles of $OH^-$ ions in $Ba(OH)_2$ , $n_2=2\times (0.15\times 60)=18$

( we multiply by 2 because 1 molecule of $Ba(OH)_2$ gives two molecule of $OH^-$ ions )

Therefore , total number of $OH^-$ ions are :

$n=n_1+n_2\\n=10+18\\n=28$

Also , total volume of solution , V = 40 + 60 = 100 ml .

Now , morality of $OH^-$ is number of moles of $OH^-$ divide by volume :

$M=\dfrac{28}{100}\\\\M=0.28 \ M$

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Molarity

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