1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76
The molar mass of CO2 is 28 grams per mole. If you are
given 2.1 moles of CO2 and you are required to find the grams of this compound,
simply multiply 28 grams per mole of CO2 by 2.1 moles of CO2 and you will get 58.8
grams of CO2.
Answer:
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