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Chemistry Calculations for pH and ppm

This document contains the answers to several chemistry questions. [1] It discusses using pH indicators that are responsive to various metal ions and can complex with EDTA near the equivalence point of a titration. [2] It shows calculations to determine the ppm of CaCO3 and MgCO3 in a sample by titrating with EDTA and measuring the pH. [3] It lists the chemical formulas of Na-EDTA and Eriochrome Black T indicator.

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Al Faricky
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39 views3 pages

Chemistry Calculations for pH and ppm

This document contains the answers to several chemistry questions. [1] It discusses using pH indicators that are responsive to various metal ions and can complex with EDTA near the equivalence point of a titration. [2] It shows calculations to determine the ppm of CaCO3 and MgCO3 in a sample by titrating with EDTA and measuring the pH. [3] It lists the chemical formulas of Na-EDTA and Eriochrome Black T indicator.

Uploaded by

Al Faricky
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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JAWABAN PERTANYAAN

1. Pemilihan indikator terkait dengan penggunaaan pH,


karena dibutuhkan indikator yang dapat renponsif
terhadap pMg, pCa, pCu, dan p yang lainnya, dan
karena indikator tersebut harus dapat melepaskan ion
metal pada EDTA apda sebuah nilai pM yang amat
dekat dengan nilai pM pada titik ekivalen.
2. Diketahui : V air = 100mL

V EDTA = 15,28mL
M EDTA = 0,0106M
pH =10
Ditanya : ppm CaCO3 dan MgCO3 ?
Jawab : mmol air = mmol EDTA
= 15,28 x 0,01016
= 0,1552 mmol
Mg CaCO3 = mmol air x Mr CaCO3
= 0,1552 x 100
= 15,5200 mg
ppm = 155,2000 mg/L
mmol air = mmol EDTA
= 10,43 x 0,01016
= 0,1059 mmol
𝑔
mmol MgCO3 = 𝑀𝑟

g = 0,1059 x 84
= 8,90148 mg
ppm = 89,0148 mg/L
3. Rumus kimia Na-EDTA:
Rumus kimia Eriochrome Black T (EBT):

4. CaCO3 + 2HCl  CaCl2 + CO2 + H2O


Massa CaCO3 = 0,0811 g
= 81,1 mg
Volume CaCO3= 100 mL = 0,1 L
𝑚𝑔
Konsentrasi CaCO3 dalam ppm = 𝐿 ppm = 811 ppm
5. NH4OH + NH4Cl→ buffer
𝐾𝑏 𝑏𝑎𝑠𝑎
[OH-] = 𝑔𝑎𝑟𝑎𝑚

𝐾𝑏 𝑁𝐻4 𝑂𝐻
= 𝑁𝐻4
1,8 𝑥 10−5 𝑥 [𝑁𝐻4 𝑂𝐻]
= [𝑁𝐻4 ]

pH = 14 – pOH
pH= 14 – pOH
pOH = 4
pOH= -log [OH-]
4= -log [OH-]
[OH-]= 10-4
1,8 𝑥 10−5 𝑥 [𝑁𝐻4 𝑂𝐻]
10-4= [𝑁𝐻4 ]

[𝑁𝐻4 𝑂𝐻] 1,8 𝑥 10−5


=
[𝑁𝐻4 ] 10−4
𝑔
[𝑁𝐻4 𝑂𝐻] =
𝑀𝑟 .𝑉
-5 𝑔
1,8 x 10 = 35 .𝑉

6,3 x 10-4 . V= g
Misal volume NH4OH =1L
Maka massa = 6,3 x 10-4 g
𝑔
[NH4OH] = 𝑀𝑟 .𝑉
𝑔
[NH4] = 𝑀𝑟 .𝑉

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