Please help ASAP
17.8 mL of HNO3 is neutralized in a titration by 24.7 mL of a 0.299 M Sr(OH)2 solution. Calculate the pH of the unknown acid.

0.1

Explanation:

We must first put down the equation of the reaction in order to guide our solution of the question.

2HNO3(aq) + Sr(OH)2(aq) > Sr(NO3)2(aq) + 2H2O(l)

Now from the question, the following were given;

Concentration of acid CA=

Concentration of base CB= 0.299M

Volume of acid VA= 17.8ml

Volume of base VB= 24.7ml

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB= CBVBNA

CA= CBVBNA/VANB

SUBSTITUTING VALUES;

CA= 0.299 × 24.7 ×2 / 17.8×1

CA= 0.8298 M

But;

pH= -log[H^+]

[H^+] = 0.8298 M

pH= -log[0.8298 M]

pH= 0.1

the awnser is group 12

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