Chemistry, 24.05.2020 01:58, oofoofoof1

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

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answered: dondre54

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

answered: Guest

I'm not sure but i will try to find out

answered: Guest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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