Question

When 100cc of 1m H2So4 is mixed with 20cc of 5m NaoH the resulting solution is.?

options below

a) acidic
b) slightly alkaline
c) strongly alkaline
d) neutrl​

Answer 1

Answer:

b) is the ans wer

Explanation:

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Answer 2

Answer:

The resulting solution is $\mathrm{Na}_{2} \mathrm{SO}_{4}$ of molarity $0.006 \mathrm{M} .$

The correct answer is Option A

Explanation

The balanced chemical reaction is;

$\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{H}_{2} \mathrm{O}$

We know that $1 c c=1 \mathrm{ml}$

Moles of $\mathrm{H}_{2} \mathrm{SO}_{4} in 100 \mathrm{ml} of 1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} solution =\left(\frac{1 \mathrm{~mol}}{1000 \mathrm{ml}}\right) * 100 \mathrm{ml}=0.1 \mathrm{~mol}$

Moles of $\mathrm{NaOH} in 20 \mathrm{ml} of 5 \mathrm{M} \mathrm{NaOH} solution =\left(\frac{5 \mathrm{~mol}}{1000 \mathrm{ml}}\right) * 20 \mathrm{ml}=0.1 \mathrm{~mol}$

But we know from a balanced equation, $2 \mathrm{~mol} of \mathrm{NaOH} reacts with 1 \mathrm{~mol} of \mathrm{H}_{2} \mathrm{SO}_{4}$

$\therefore$ for given $0.5$ moles of$\mathrm{H}_{2} \mathrm{SO}_{4}$ remain unreacted.

The final volume of solution$=100+20=120 \mathrm{ml}$

Molarity of resulting solution of $\mathrm{Na}_{2} \mathrm{SO}_{4}=\left(\frac{0.05 \mathrm{~mol}}{1000 \mathrm{ml}}\right) *(120)=0.006 \mathrm{M}$

Thus Meq of acid is left and therefore pH < 7, so the resulting mixture will be acidic.