The temperature for each solution is carried out at approximately 297 K where Kw
ID: 558819 • Letter: T
Question
The temperature for each solution is carried out at approximately 297 K where Kw 1.00x 10-14 Part A 0.55 g of hydrogen chloride (HCl) is dissolved in water to make 7.0L of solution. What is the pH of the resulting hydrochloric acid solution? Express the pH numerically to two decimal places. Hints pH- Submit My Answers Give Up Part B 0.45 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 60 L of solution. What is the pH of this solution? Express the pH numerically to two decimal places Hints pH- Submit My Answers Give UpExplanation / Answer
A)
Molar mass of HCl = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol
mass of HCl = 0.55 g
we have below equation to be used:
number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(0.55 g)/(36.458 g/mol)
= 1.509*10^-2 mol
volume , V = 7.0 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 1.509*10^-2/7
= 2.155*10^-3 M
So,
[H+] = 2.155*10^-3 M
we have below equation to be used:
pH = -log [H+]
= -log (2.155*10^-3)
= 2.67
Answer: 2.67
B)
Molar mass of NaOH = 1*MM(Na) + 1*MM(O) + 1*MM(H)
= 1*22.99 + 1*16.0 + 1*1.008
= 39.998 g/mol
mass of NaOH = 0.45 g
we have below equation to be used:
number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(0.45 g)/(39.998 g/mol)
= 1.125*10^-2 mol
volume , V = 6.0 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 1.125*10^-2/6
= 1.875*10^-3 M
[OH-] = 1.875*10^-3 M
we have below equation to be used:
pOH = -log [OH-]
= -log (1.875*10^-3)
= 2.73
we have below equation to be used:
PH = 14 - pOH
= 14 - 2.73
= 11.27
Answer: 11.27
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