17.8 mL of HNO3 is neutralized in a titration by 24.7 mL of a 0.299 M Sr(OH)2 solution. Find the pH of the acid

0.08

Explanation:

First, we begin by writing the balanced equation for the reaction. This is illustrated below:

2HNO3 + Sr(OH)2 → Sr(NO3)2 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 2

The mole ratio of the base (nB) = 1

Next, the data obtained from the. This includes the following:

Volume of acid (Va) = 17.8 mL

Volume of base (Vb) = 24.7 mL

Molarity of base (Mb) = 0.299 M

Molarity of acid (Ma) =..?

pH =..?

Next, we shall determine the molarity of the acid. This can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 17.8/0.299 x 24.7 = 2/1

Cross multiply to express in linear form

Ma x 17.8 = 0.299 x 24.7 x 2

Divide both side by 17.8

Ma = (0.299 x 24.7 x 2) /17.8

Ma = 0.830M.

Therefore, the molarity of the acid, HNO3 is 0.830M

Finally, we can obtain the pH of the acid as follow:

We shall obtain the concentration of the hydrogen ion, [H+]. This is illustrated below:

HNO3 —> H+ + NO3-

From the balanced equation above,

1 mole of HNO3 produced 1 mole of H+.

Therefore, 0.830M HNO3 will also produce 0.830M H+.

Therefore, the concentration of the hydrogen ion, [H+] is 0.830M

pH = –Log [H+]

[H+] = 0.830M

pH = –Log 0.830M

pH = 0.08

Therefore, the pH of the acid is 0.08

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