0.80 g of hydrogen chloride (HCL) is dissolved in water to make 6.5 L of solutio

Question

0.80 g of hydrogen chloride (HCL) is dissolved in water to make 6.5 L of solution. What is the pH of the resulting hydrochloric acid solution? Express the pH numerically to two decimal places. 0.25 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 5.0 L of solution. What is the pH of this solution? Express the pH numerically to two decimal places. At a certain temperature, the pH of a neutral solution is 7.38. What is the value of K_w at that temperature? Express your answer numerically using two significant figures. If K_b for NX_3 is 9.0 times 10^-6, what is the pOH of a 0.175 M aqueous solution of NX_3 ? Express your answer numerically. If K_b for NX3 is 9.0 times 10^-6, what is the percent ionization of a 0.325 M aqueous solution of NX Express your answer numerically to three significant figures. If K_b for NX_3 is 9.0 times 10^-6, what is the the pK_a for the following reaction? Express your answer numerically to two decimal places.

Explanation / Answer

PART A

Supposing the HCl ionizes completely because it is a strong acid:

Molar mass of HCl = 36.46 g/mol

0.8 g of HCL = (0.8/ 36.46) mol = 0.022 mol

So, 0.022 mol in 6.5 L =0.022 mol/6.5 L = 0.0034 M

So, [H+] = 0.0034 M

pH = - log [H+] = - log [0.0034] = 2.47

PART B

Molar mass of NaOH = 40 g/mol

0.25 g of NaOH = (0.25/ 40) mol = 0.00625 mol

So, 0.00625 mol in 5 L =0.00625 mol/5 L = 0.00125 M

Which is concentration of [OH-]

So, [OH-] = 0.00125 M

pOH = - log [OH-] = - log [0.00125] = 2.90

pH = 14 - pOH = 14 – 2.9 = 11.1

PART C:

It is very important to know that neutral pH changes with temperature.

The key part in this problem is that the solution is neutral! In a neutral solution the pH and pOH are equal and therefore so are their concentrations.

Kw= [H+] x [OH-] where [H+] and [OH-] represent the concentrations of those ions.

pH =7.38

-log[H+] = 7.38

[H+]=10-7.38 = 4.17 x 10-8

Since it is a neutral solution [H+]=[OH-],

Kw[H+] x [OH-]

Kw=(4.17 x 10-8)2 = 1.74 x 10-15

PART A (Right side)

NX3 + H2O <----> NHX3+ + OH-

Kb = 9 x 10-6 = x2 / 0.175-x

x2 + (9 x 10-6)x –(1.575 x 10-6) = 0

Solving quadratic equation, we get x = 1.25 x 10-3

So, x = [OH-]=1.25 x 10-3 M
pOH = -log (1.25 x 10-3) = 2.9

Related Questions

Navigate

• Browse (All)
• Browse
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.