#22 The temperature for each solution is carried out at approximately 297 K wher

Question

#22 The temperature for each solution is carried out at approximately 297 K where Kw=1.00×1014.

Part A

0.20 g of hydrogen chloride (HCl) is dissolved in water to make 4.0 L  of solution. What is the pH of the resulting hydrochloric acid solution?

Express the pH numerically to two decimal places.

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Part B

0.55g of sodium hydroxide (NaOH) pellets are dissolved in water to make 5.5 L of solution. What is the pH of this solution?

Express the pH numerically to two decimal places.

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Explanation / Answer

A)

Molar mass of HCl,
MM = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol


mass(HCl)= 0.20 g

use:
number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(0.2 g)/(36.458 g/mol)
= 5.486*10^-3 mol
volume , V = 4.0 L


use:
Molarity,
M = number of mol / volume in L
= 5.486*10^-3/4
= 1.371*10^-3 M
So,
[H+] = [HCl] = 1.371*10^-3 M

use:
pH = -log [H+]
= -log (1.371*10^-3)
= 2.86
Answer: 2.86

B)

Molar mass of NaOH,
MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)
= 1*22.99 + 1*16.0 + 1*1.008
= 39.998 g/mol


mass(NaOH)= 0.55 g

use:
number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(0.55 g)/(39.998 g/mol)
= 1.375*10^-2 mol
volume , V = 4.0 L


use:
Molarity,
M = number of mol / volume in L
= 1.375*10^-2/4
= 3.438*10^-3 M

SO,
[OH-] = [NaOH] = 3.438*10^-3 M

use:
pOH = -log [OH-]
= -log (3.438*10^-3)
= 2.46


use:
PH = 14 - pOH
= 14 - 2.46
= 11.54
Answer: 11.54

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