Yl12_lah - 1. Arvutage pH ja pOH järgmistes tugevate...

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Unformatted text preview: 1. Arvutage pH ja pOH järgmistes tugevate hapete või aluste vesilahustes: a) 0.11M HCl; b) 0.0092M Ba(OH) 2 ; c) 27.9 mg NaOH, mis on lahustatud 0.35 l vees; d) 75.0 ml 3.5*10-4 M HBr lahus peale lahjendamist 0.250 liitrini Lahendus : a) pH=-log[H 3 O + ]=-log c HCl =-log0.11 b) pH=14-log[OH- ]=14-log (2*c Ba(OH)2 )=14-log(2*0.0092) c) c NaOH =27.9*10-3 /(40*0.35)=0.002M; pH=14-log[OH- ]=14-log (c NaOH )=14-log(0.002) d) c HBr =(0.075*3.5*10-4 )/0.25=1.05*10-4 ; pH=-log[H 3 O + ]=-log c HBr =-log 1.05*10-4 Vastused : a) 0.96 ja 13.04; b) 12.26 ja 1.74; c) 11.30 ja 2.70; d) 3.98 ja 10.02 2. Bensoehappe, C 6 H 5 COOH, 0.110M lahuses on dissotsieerunud 2.4% happest. Arvutage lahuse pH ja bensoehappe pK! Lahendus: [H 3 O + ]=[C 6 H 5 COO- ]=0.024*0.110=0.00264M; pH=-log[H 3 O + ]=-log 0.00264 [C 6 H 5 COOH]=0.024-0.00264=0.107; K=(0.00264*0.00264)/0.107=6.5*10-5 ; pK=-logK=-log 6.5*10-5 ; Vastused : 2.58 ja 4.19 3. Arvutage 6.55*10-8 M HClO 4 lahuse pH!...
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