hi~o(*∩_∩*)o~everybody

Hi~~o(*∩_∩*)o

On friday we corrected page 9 the problems 1 to 9 . . .
Here the answers : . . .

1. NaOH + HNO3 NaNO3 + H2O
mol NaOH = 0.20 mol /1 * 1 = 0.20 mol

2. HCl + KOH H2O + KCl
x (0.45 mol/L) = 25.0 mL (1.00 mol/L)
x = 56 mL

3. a) HCl + NaOH H2O + NaCl
mol HCl = 0.20 mol / 1 * 1 = 0.20 mol

b) 2HCl + Ca(OH)2 2H2O + CaCl2
mol HCl = 0.10 mol /1 * 2 = 0.20 mol

4. HCl + NaOH H2O + NaCl
28.0 mL / 1000 = 0.028 L
mol HCl = 0.028 L (1.00 mol/L) = 0.028 mol
20.0mL / 1000 = 0.020 L
C NaOH = 0.028 mol / 0.020 L = 1.40 mol/L

5. End point - when performing a tritration the standard is added until one drop permanently change the colour of indicator.

Equivalence point - As the two solution are mixed, the system will reach a point at which the acid and base will have neutralized one another.
- The point at which the number of mole of acid are equivalent to the number of moles of base.

6. C NaOH = x (43.0 mL) = 32.0 mL (0.100 mol/L)
x = 0.0744 mol/L

7. C NaOH = x (40.0 mL) = 25.0 mL (0.120 mol/L)
x = 0.075mol/L

8. C KOH = x (50.0 mL) = 30.0 mL (0.250mol/L)
x = 0.15 mol/L

9. mol of acid = 25.0 mL / 1000 * 1.44 mol/L = 0.036 mol
mol of base = 36.0 mL / 1000 * 2.00 mol / L =0.072 mol

Acid must be diprotic ( the base has one OH ion molecules and twice as many moles of base are required to neutralize the acid)

And don't forget to do the worksheet~~

The next scribe Rawra

o(*∩_∩*)o