7/24/2019 Perhitungan Anor 1 Prak 3

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PERHITUNGAN:

1 Molaritas Asam Oksalat

M=massaH

2C

2O

4.2H

2O

Mr H2C2O4.2H2O x

1000mL

V lar . as. oksalatx 1L

M=0,63 g

126 g

mol

x1000mL

50mL x1L

M=0,1mo /L

2 Standarisasi NaOH 0,1 M

Reaksi : 2 NaOH + H2C2O4Na2C2O4+ 2 H2O

MNaOHx Vratarata NaOH=2

(MH2C2 O 4x

VH

2 C2 O4)

MNaOH=2(MH

2C

2O

4

x VH2

C2O

4

)

Vratarata NaOH

MNaOH=2(0,1M x 10mL)

21,7mL

MNaOH=0,09M

3 Penentuan [Na+ den!an "antuan Resin Penukar #ation

massa Na= Ar NaMr NaCl

xmassa NaCl mulamula

massa Na=23

58,5x0,2 gram

massa Na=0,08gram

mol Na=massa Na

Ar Na

mol Na= 0,08g23g /mol

=3,48x103 mol=3,48mmol

Reaksi $ertukaran ion

R%H + Na+R%Na + H+

H++ OH%H2O

&ari reaksi diketa'ui (a')a mol OH%: mol H+: mol Na+ * 1 :1 :1mol Na+* mol H+ * mol OH%* NaOH MNaOH

7/24/2019 Perhitungan Anor 1 Prak 3

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* -,.- m/ 0,1 mmolml

* 0,3 mmol

er5adi $en!en6eran . kali ter'ada$ e77luent:mol Na+* . 0,3 mmol * -,-. mmol

Efisiensi=mol N a yang dipertukarkanmol Na awal x100

Efisiensi=3,35mmol

3,48mmolx 100

Efisiensi=96,3

4 Penentuan Hasik #ali #elarutan, #s$ MC2O48M* Ca, M!, dan "a9Reaksi $ertukaran ion

2 R%H + M2+R2%M + 2 H+

2 H+

+ 2 OH%

2 H2O&ari reaksi diketa'ui (a')a mol OH%: mol H+: mol M+ * 2 :2 :1

a Penentuan #s$ M!C2O4Mol OH%* NaOH MNaOHMol OH%* 0,2- m/ 0,0 M * 0,021 mmol

Mol M!2+*1

2x mol OH

Mol M!2+*1

2x0,021mmol=0,011mmol

er5adi $en!en6eran . kali $ada e77luent se'in!!a,

Mol M!2+ * . 0,011mmol

* 0,0.. mmol

[M!2+ *0,055mmol

10mL =5,5x103M

#s$ * [M!2+[ C

2O

4 2%

#s$ * S2

#s$ * (5,5x103 )2=3,03x 105

( Penentuan #s$ CaC2O4

Mol OH%* NaOH MNaOH

Mol OH%* 0,23 m/ 0,0 M * 0,024 mmol

7/24/2019 Perhitungan Anor 1 Prak 3

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Mol M!2+*1

2x mol OH

Mol M!2+*1

2x0,024mmol=0,012mmol

er5adi $en!en6eran . kali $ada e77luent se'in!!a,

Mol M!2+ * . 0,012mmol * 0,0 mmol

[M!2+ *0,055mmol

10mL =6,0x 103M

#s$ * [M!2+[ C

2O

4 2%

#s$ * S2

#s$ * (6,0x103 )2=3,6x105

6 Penentuan #s$ "aC2O4

Mol OH%* NaOH MNaOH

Mol OH%* 0,2 m/ 0,0 M * 0,01 mmol

Mol M!2+*1

2x mol OH

Mol M!2+*1

2x0,018mmol=0,009mmol

er5adi $en!en6eran . kali $ada e77luent se'in!!a,

Mol M!2+ * . 0,009mmol

* 0,04. mmol

[M!2+ *0,055mmol

10mL =4,5x103M

#s$ * [M!2+[ C

2O

4 2%

#s$ * S2

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#s$ * (4,5x 103 )2=2,03x105