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Samenvatting Antwoorden afsluiting H7 4 HAVO scheikunde chemie overal
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Antwoorden afsluiting H7 4 HAVO scheikunde chemie overal
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AfstuitingEO
a Bij 0,45 M zoutzuur geldt [H.l = 0,45 mol L-l.
pH = -log 0,45 = 0,35
b De molaire massa van HNO, is 63,013 I mo¡-t,
31.5 s HNO^ a 31'5 = 0,500 mol
' 63,013
lH'l = 0,500 mol L-l
pH = -log 0,500 = 0,301
ø
a Door verdunnen wordt de oplossing minder zuur. De [H.] wordt dus kleiner.
b De [H.] wordt kleiner. De pH wordt dus groter.
c Zie Binas tabel 52 A.
Als methyloranje oranjegeel kleurt, geldt: pH > 4,4.
Als broomkresolgroen groen kleurt, geldt: 3,8 < pH < 5,4.
De pH ligt dus tussen 4,4 en 5,4.
E
a De blauwe kleur wijst op pH < 5, het regenwater is dus zuur.
b De rode kleur wijst op een pH > 6,5, het mengsel is dus basisch.
c pH = 4,9 -+ [H*] - 10-pH = 10-4,e = 1.10-5 mol L-l
tr
a De molaire massa van picrinezuur is 229,1I g mol-l.
1'4
or5/ 22g,rr4 = 6,1'10-3
i,4 s C.H^N.O.4 mol
opgeloste stof 6'1- 1Q-] mol
molariteit - mol -
volume oplossing 0,100 L
= 6,r.!o-zmol L-l
b [H.] = 0,85 x 6,1.10-2 = 5,2'10-2 mol L-1
c pH = -log 5,2.10-2 = I,28 (Gegevens hebben twee significante cijfers, er zijn dus twee
decimalen in de pH.)
d C6H3N307 (s)-+ C.H.N.O, (B)
tr*
a 1 0H- reageert met 1 H*, er zijn dus 6,64'10 4 mol Hrionen.
a
b 1 mol citroenzuur geeft 3 mol H', er is 1/3 x 6,64.10-a = 2,2I't0 mol citroenzuur.
c De molaire massa van C6HB07 is 192,L2 g mol-l.
2,21.10-4 mol CuHrO, !2,2I'I0-4 x I92,I2 = 4,25'10-2 E
o'?';l9t x
massapercentage = 2,007
100% = 2,r20/o
tr
a Extraheren en filtreren
b [H'] = 0,2 mol L-l
pH = -log 0,2 = 0,7
c Azijnzuur is een zwak zuur en geeft dus bij dezelfde concentratie minder H'ionen.
@ Noordhoff Uitgevers bv Zuren | 101
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