OEFENTOETS Zuren en basen 5 VWO 1. 2. 3. 4. 5. 6. 8.
OEFENTOETS Zuren en basen 5 VWO 1. 2. 3. 4. 5. 6. 8.
OEFENTOETS Zuren en basen 5 VWO 1. 2. 3. 4. 5. 6. 8.
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16 [H 3O + ] = 10 -2,70 = 2,010 -3 M.<br />
Perc<strong>en</strong>tage ionisatie = [H 3O + ] / [azijnzuur] × 100% = 2,010 -3 / 0,10 ×<br />
100 = 2,0%.<br />
17 Kalkwater bevat de base OH - .<br />
CO 2 in water wordt het zwakke zuur H 2CO <strong>3.</strong><br />
Bij de reactie van H2CO3 met OH - 2-<br />
ontstaan o.a. ion<strong>en</strong> CO3 die met<br />
Ca 2+ uit kalkwater e<strong>en</strong> neerslag vorm<strong>en</strong>.<br />
H2CO3 + Ca 2+ + 2 OH - [Ca 2+ 2-<br />
+ CO3 ] + 2 H2O<br />
CO 2(aq) + Ca 2+ + 2 OH - H 2O + CaCO 3(s)<br />
18 MgCO 3 + 2 H 3O + [ Mg 2+ + 2 H 2O + ‘H 2CO 3’ ] Mg 2+ + 3 H 2O + CO 2(g).<br />
19 CaH 2 + H 2O Ca 2+ + 2 OH - + H 2(g).<br />
44,0 mg<br />
÷ 42,0<br />
1,05 mmol → (1:2) 2,10 mmol in 200 mL → [OH - ] = 0,0105 M.<br />
pOH = - log 0,0105 = 1,98<br />
pH = 14,00 – 1,98 = 12,0<strong>2.</strong><br />
20 4,77% van 1,029 = 0,0491 kg = 49,1 gram H 2SO <strong>4.</strong><br />
÷ 98,08 = 5,0010 -1 mol H 2SO 4 per liter<br />
× 2 = 1,00 mol H 3O + per liter.<br />
Hiervan 50,0 mL verdunn<strong>en</strong> tot 1000 mL → factor 20<br />
Dus [H 3O + ] = 1,00/ 20 = 5,0010 -2 M.<br />
pH = - log (5,0010 -2 ) = 1,<strong>3.</strong>
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