1. Calculate the pH and pOH of the solutions given below. (log5 =0.7, log8 = 0.9) a. A solution with a [H + ] of 1x10 -3 M. b. A solution with a [OH - ] of 1x10 -3 M. c. A solution with a [H + ] of 5x10 -7 M. d. A solution with a [OH - ] of 8x10 -11 M.
1. 1. Calculate the pH and pOH of the solutions given below. (log5 =0.7, log8 = 0.9) a. A solution with a [H + ] of 1x10 -3 M. b. A solution with a [OH - ] of 1x10 -3 M. c. A solution with a [H + ] of 5x10 -7 M. d. A solution with a [OH - ] of 8x10 -11 M.
pH = -log [H⁺]
pOH = -log [OH⁻]
pH = 14-pOH
a) pH = 3-log 10 = 3
pOH = 14-3 = 11
b) pOH = 3-log 10 = 3
pH = 14-3 = 11
c) pH = 7-log 5 = 6,3
pOH = 14-6,3 = 7,7
d) pOH = 11-log 8 = 10,09
pH = 14-10,09 = 3,91
2. A solution of H2SO4 with a molal concentration of 8.010 m has adensity of 1.354g/mL. What is the molar concentration of thissolution?
Answer with explanation:
You are given a solution of sulphuric acid with a molal concentration of 8.010 m and its density is 1.354 g/mL.
And we also know that the molar mass of sulphuric acid which is 98.079 g/mol.
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With the known datas, we work out the molar concentration in mol/L.
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The relation between molarity, molality and density is described within the following equation.
[tex]\frac{1}{m} = \frac{d}{M}-\frac{Mr}{1000}[/tex]
Where,
Density (d) = 1.354 g/mL
Molecular mass of solute (Mr) = 98.079 g/mol
Molality (m) = 8.010 m
Molarity (M) = ... mol/L
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So then,
[tex]\frac{1}{8.010\ m} = \frac{1.354\ g/mL}{M}-\frac{98.079\ g/mol}{1000}[/tex]
[tex]0.1248\ m = \frac{1.354\ g/mL}{M}-0.098079\ g/mol[/tex]
[tex]\frac{1.354\ g/mL}{M} = 0.098079\ g/mol+ 0.1248\ m[/tex]
[tex]\frac{1.354\ g/mL}{M} = 0.222879\ g/mmol[/tex]
[tex]M = \frac{1.354\ g/mL}{0.222879\ g/mmol}[/tex]
[tex]M = 6.075\ mol/L[/tex]
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Therefore, the molar concentration of said acid solution is 6.075 mol/L.
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~I hope this helps you~
3. Hitunglah pH larutan 500mL HCN 0, 1 M yang dicampurkan dengan 500mL KOH 0,1 M! (Ka HCN = 10^-6)
* Mencari mol
mol HCN = M • V
mol HCN = 0,1 • 500
mol HCN = 50 mmol
mol KOH = M • V
mol KOH = 0,1 • 500
mol KOH = 50 mmol
* Mencari mol anion garam
Persamaan reaksi :
HCN + KOH → KCN + H₂O
a 50 50
r 50 50
__________________________
s – – 50 mmol
mol KCN = 50 mmol
KCN ⇆ K⁺ + CN⁻
50 50 50
mol CN⁻ = 50 mmol
* Mencari konsentrasi anion garam
M = mol/Volume total
M = 50/(500 + 500)
M = 50/1000
M = 5 x 10⁻² M
* Mencari [OH⁻]
[OH⁻] = √(Kw/Ka • M anion)
[OH⁻] = √(10⁻¹⁴/10⁻⁶ • 5 x 10⁻²)
[OH⁻] = √(5 x 10⁻¹⁰)
[OH⁻] = 2,24 x 10⁻⁵
* Mencari pOH
pOH = -log [OH⁻]
pOH = -log [2,24 x 10⁻⁵]
pOH = 5 - log 2,24
* Mencari pH
pH = 14 - pOH
pH = 14 - (5 - log 2,24)
pH = 9 + log 2,24
Jadi, pH akhir larutan adalah 9 + log 2,24
4. Untuk membuat 500ml larutan KOH 0,2M diperlukan kristal KOH murni sebanyak (M, KOH = 56)...
V = 500 mL
M = 0,2 molar
Mr = 56
wt = ....?
M = wt/Mr x 1000/V
0,2 = wt/56 x 1000/500
wt x 2 = 56 x 0,2
wt = 11,2/2
wt = 5,6 gram
5. The solution of 3(m – 1) + 5 = m + 10 is m =
Jawab:
m = 4
Penjelasan dengan langkah-langkah:
3(m-1) + 5 = m + 10
3m - 3 + 5 = m + 10
3m + 2 = m + 10
3m - m = 10 - 2
2m = 8
m = 4
Hope this helps!
6. What is the molarity of a solution containing 4 moles of KCl in 2.5 L of solution?
Penjelasan:
dik : n KCl = 4 mol
v = 2,5 L
dit : M = ?
jb :
M = n / v
= 4 mol / 2,5 L
= 1,6 mol/L
semoga membantu :)
7. 4. Find the solution of x² + 8x +16 0!
Jawab:
x² + 8x +16 = 0
(x + 4)^2 = 0
x = - 4
8. A 20 ml solution of 0,0512 M oxalic acid was used to standardise an unknown solution of potassium permanganate. An average of 21,24 ml of permanganate was required. Calculate the concentration of the potassium permanganate solution
Jawaban:
0.0964 M
Penjelasan:
H2C2O4 = KMnO4
V1 × N1 = V2 × N2
20 × 0.0512 × 2 = 21.24 × M × 1
[tex]M = \frac{20 × 0.0512 × 2 }{21.24} \\ M = 0.0964 [/tex]
9. Untuk membuat 500ml larutan koh 0.2 m diperlukan kristal koh murni sebanyak mr koh=56
volume = 500 ml
Molaritas = 0,2 M
mol = molaritas x volume
= 500 x 0,2
= 100 mmol
Mr KOH = 56
massa = mol x mr
= 100 x 56
= 5600 mg
= 5,6 gram
10. ph dari campuran 500ml ch3cooh 0,2 m ka 10-5 g 500ml koh 0,2 m adalah
diket: V asam lemah=500 ml, M asam lemah 0,2 M, V basa kuat=500 ml, M basa kuat=0,2 M, Ka=10^-5 Vtotal=1000 ml
ditanya:PH campuran?
penyelesaian
mol asam=M×V=0,2×500=100 mmol
mol basa=M×V=0,2×500=100 mmol
karena keduanya sama maka akan habis bereaksi
M garam=mol/Vtot=100/1000=0,1
OH= √Kw/Ka×M garam
= √10^-14/10^-5×0,1
= √10^-9×0,1
= √10^-10
=10^-5
PH =PKw-POH
=14-(-log10^-5)
=14-5
=9
11. Berapakah massa koh untuk membuat 500ml koh 0,4 m jika mr koh=56
mol KOH = M x V
= 0,4 M x 0,5 L = 0,2 mol
massa = mol x Mr = 0,2 x 56 = 11,2 gram
12. untuk membuat 500ml larutan KOH 0,2 M diperlukan kristal KOH murni sebanyak ...
Mr KOH = 56
mol KOH (n) = M x volume (liter)
n = 0,2 x 0,5 liter
n = 0,1 mol
massa KOH (w) = mol x Mr
w = 0,1 x 56
w = 5,6 grMR koh=56
ML=500 x 0.2=100MML=0.1ML
ML=massa/MR Koh
Massa=ML x Mr Koh
=0,1 x 56 =56 gram
Jadi untuk membuat 500ml larutan diperlukan kristal KOH murni sebanyak 56G
#Maaf Kalau Salah
13. Hitung pH dari larutan KOH 0,01 M 500mL
Jawaban:
pH larutan KOH 0,01 M sebanyak 500 mL adalah 12. Langkah penyelesaian
Menghitung valensi basa Menghitung [OH⁻]Menghitung pOH Menghitung pHPenjelasan:
Langkah I : Menghitung valensi basa
KOH ⇔ K⁺ + OH⁻
Karena jumlah OH⁻ yang terionisasi 1 maka jumlah valensi basa adalah 1
Langkah II : Menghitung [OH⁻]
[OH⁻] = molaritas basa x valensi basa
= 0,01 x 1
= 0,01
= 10⁻²
Langkah III : Menghitung pOH
pOH = - log [OH⁻]
= - log (10⁻²)
= 2
Langkah IV : Menghitung pH
pH = 14 - pOH
= 14 - 2
= 12
Pelajari lebih lanjut tentang materi menghitung pH KOH, pada https://brainly.co.id/tugas/37876012
#BelajarBersamaBrainly
14. seorang siswa diminta menyediakan 500mL larutan KOH 0,2M. massa KOH yg diperlukan sebanyak.....(M,KOH=56)
mol KOH = 0,5 L.0,2 M = 0,1 mol
massa KOH = 0,1 mol. 56 gram/mol = 5,6 gram
15. untuk membuat 500ML larutan KOH 0,2 M diperlukan kristal KOH murni sebanyak... (Mr KOH= 56)
1. dicari mol KOH dulu, caranya : Volume x konsentrasi
mol= 500 x 0,2 = 100 mmol = 100 x [tex] 10^{-3} [/tex]
=0,1 mol
2. dicari massa kristal KOH
mol= massa/Mr KOH
massa = mol x Mr KOH
= 0,1 x 56 = 5,6 gram kristal KOH
Volume x konsentrasi
mol= 500 x 0,2 = 100 mmol = 100 x
=0,1 mol
2. dicari massa kristal KOH
mol= massa/Mr KOH
massa = mol x Mr KOH
= 0,1 x 56 = 5,6 gram kristal KOH
16. find the solution set of |x-2|+1 = 0 a. -6 b.-3 c.0 d. no solution e. 6
|x - 2| = -1
x - 2 = -1
x = 1
atau
-(x - 2) = -1
-x + 2 = -1
3 = x
17. Water is added to 25.0 ml of a 0.866 m kno3 solution until the volume of the solution is exactly 500 ml. what is the concentration of the fi nal solution?
Jawaban:
Penjelasan:
[tex]M_{KNO_{3}}= 0,866 \ M = \frac{n}{V_{i}} \\\\V_{f}= 500 \ mL\\\\\delta V=25 \ mL\\\\V_{i}=V_{f}-\delta V\\\\V_{i}=500 \ mL-25 \ mL = 475 \ mL\\\\M_{f \ KNO_{3}}=M_{KNO_{3}}\times \frac{V_{i}}{V_{f}} =\\\\M_{f \ KNO_{3}}=0,866 \ M \times \frac{475 \ mL}{500 \ mL} =\\\\M_{f \ KNO_{3}}=0,866 \ M \times \frac{475 \ mL}{500 \ mL} = 0,8227 \ M[/tex]
18. If p = 2 is a solution of the equation 2p - hp +6=0, find the value of h. Hence, find the other solution for the equation.
remember p = 2
Then,
2p² - hp + 6 = 0
2(2)² - h(2) + 6 = 0
2(4) - 2h + 6 = 0
8 + 6 - 2h = 0
14 - 2h = 0
2h = 14
h = 7
so, we have solution h is a 7
19. Which definition of concentration given below best describes the concept expressed in “in a solution of 10g of salt and 70ml of water, the concentration of the solution by percent mass is 12,5%2?” jawabannya 1.The amount of water in the solution 2 A word denoting 70g 3 The mass of the solution 4 The amount of a component (solute) present. 5 The amount of solution
Jumlah 1. air dalam larutan 2 A 70g kata yang menunjukkan 3 Massa solusi 4 Jumlah komponen ( zat terlarut ) hadir . 5 Jumlah solusi
20. hitunglah pH larutan 500mL HCN 0,1 M yang di campurkan dengan 500mL KOH 0,1 M
Mencari mol
mol HCN = M • V
mol HCN = 0,1 • 500
mol HCN = 50 mmol
mol KOH = M • V
mol KOH = 0,1 • 500
mol KOH = 50 mmol
* Mencari mol anion garam
Persamaan reaksi :
HCN + KOH → KCN + H₂O
a 50 50
r 50 50
__________________________
s – – 50 mmol
mol KCN = 50 mmol
KCN ⇆ K⁺ + CN⁻
50 50 50
mol CN⁻ = 50 mmol
* Mencari konsentrasi anion garam
M = mol/Volume total
M = 50/(500 + 500)
M = 50/1000
M = 5 x 10⁻² M
* Mencari [OH⁻]
[OH⁻] = √(Kw/Ka • M anion)
[OH⁻] = √(10⁻¹⁴/10⁻⁶ • 5 x 10⁻²)
[OH⁻] = √(5 x 10⁻¹⁰)
[OH⁻] = 2,24 x 10⁻⁵
* Mencari pOH
pOH = -log [OH⁻]
pOH = -log [2,24 x 10⁻⁵]
pOH = 5 - log 2,24
* Mencari pH
pH = 14 - pOH
pH = 14 - (5 - log 2,24)
pH = 9 + log 2,24
Jadi, pH akhir larutan adalah 9 + log 2,24
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