The molarity of an aqueous solution of sodium hydroxide, NaOH, is determined by titration against a 0.102 M hydrochloric acid, HCl, solution HCINaOH-NaClH20 If 30.1 mL of the base are required to neutralize 21.7 mL of hydrochloric acid, what is the molarity of the sodium hydroxide solution? Molarity
Solution
1.
HCl + NaOH -------> NaCl + H2O
In the above reaction equation; HCl and NaOH reacts in 1:1 ratio.
21.7 mL of 0.102 M HCl
So, moles of HCl = 0.102 M x 0.0217 L = 0.0022134 moles
Hence, the moles of NaOH will also be equal to 0.0022134 moles
Volume of NaOH = 30.1 mL = 0.0301 L
[NaOH] = 0.0022134 mol / 0.0301 L = 0.074 M
2.
HBr + Ba(OH)2 -------> BaBr2 + 2H2O
In the above reaction equation; HBr and Ba(OH)2 reacts in 2:1 ratio.
28.0 mL of 0.0471 M Ba(OH)2
So, moles of Ba(OH)2 = 0.0471 M x 0.028 L = 0.0013188 moles
Hence, the moles of HBr = (0.0013188 x 2) moles = 0.0026376 moles
Volume of HBr = 16.9 mL = 0.0169 L
[HBr] = 0.0026376 mol / 0.0169 L = 0.156 M
3.
pOH = 6.10
pH = 14 – pOH = 14 – 6.10 = 7.90
pOH = 6.10
Or, - log [OH-] = 6.10
Or,[OH-] = 10 -6.10 = 7.94 x 10^-7
pH = 7.90
Or, - log [H3O+] = 7.90
Or,[H3O+] = 10 -7.90 = 1.26 x 10^-8
4.
pH = 7.20
pOH = 14 – pH = 14 – 7.20 = 6.80
pOH = 6.80
Or, - log [OH-] = 6.80
Or,[OH-] = 10 -6.80 = 1.58 x 10^-7
pH = 7.90
Or, - log [H3O+] = 7.20
Or,[H3O+] = 10 -7.20 = 6.31 x 10^-8
.
Call Girls in Dwarka Mor Delhi Contact Us 9654467111
NaOH Molarity by Titration vs HCl
1. The molarity of an aqueous solution of sodium hydroxide, NaOH, is determined by titration
against a 0.102 M hydrochloric acid, HCl, solution HCINaOH-NaClH20 If 30.1 mL of the base
are required to neutralize 21.7 mL of hydrochloric acid, what is the molarity of the sodium
hydroxide solution? Molarity
Solution
1.
HCl + NaOH -------> NaCl + H2O
In the above reaction equation; HCl and NaOH reacts in 1:1 ratio.
21.7 mL of 0.102 M HCl
So, moles of HCl = 0.102 M x 0.0217 L = 0.0022134 moles
Hence, the moles of NaOH will also be equal to 0.0022134 moles
Volume of NaOH = 30.1 mL = 0.0301 L
[NaOH] = 0.0022134 mol / 0.0301 L = 0.074 M
2.
HBr + Ba(OH)2 -------> BaBr2 + 2H2O
In the above reaction equation; HBr and Ba(OH)2 reacts in 2:1 ratio.
28.0 mL of 0.0471 M Ba(OH)2
So, moles of Ba(OH)2 = 0.0471 M x 0.028 L = 0.0013188 moles
Hence, the moles of HBr = (0.0013188 x 2) moles = 0.0026376 moles
Volume of HBr = 16.9 mL = 0.0169 L
[HBr] = 0.0026376 mol / 0.0169 L = 0.156 M
3.
pOH = 6.10
![The molarity of an aqueous solution of sodium hydroxide, NaOH, is determined by titration
against a 0.102 M hydrochloric acid, HCl, solution HCINaOH-NaClH20 If 30.1 mL of the base
are required to neutralize 21.7 mL of hydrochloric acid, what is the molarity of the sodium
hydroxide solution? Molarity
Solution
1.
HCl + NaOH -------> NaCl + H2O
In the above reaction equation; HCl and NaOH reacts in 1:1 ratio.
21.7 mL of 0.102 M HCl
So, moles of HCl = 0.102 M x 0.0217 L = 0.0022134 moles
Hence, the moles of NaOH will also be equal to 0.0022134 moles
Volume of NaOH = 30.1 mL = 0.0301 L
[NaOH] = 0.0022134 mol / 0.0301 L = 0.074 M
2.
HBr + Ba(OH)2 -------> BaBr2 + 2H2O
In the above reaction equation; HBr and Ba(OH)2 reacts in 2:1 ratio.
28.0 mL of 0.0471 M Ba(OH)2
So, moles of Ba(OH)2 = 0.0471 M x 0.028 L = 0.0013188 moles
Hence, the moles of HBr = (0.0013188 x 2) moles = 0.0026376 moles
Volume of HBr = 16.9 mL = 0.0169 L
[HBr] = 0.0026376 mol / 0.0169 L = 0.156 M
3.
pOH = 6.10](https://image.slidesharecdn.com/themolarityofanaqueoussolutionofsodiumhydroxide-naoh-isdete-230203020537-e3f2308d/85/The-molarity-of-an-aqueous-solution-of-sodium-hydroxide-NaOH-is-dete-docx-1-320.jpg)
![pH = 14 – pOH = 14 – 6.10 = 7.90
pOH = 6.10
Or, - log [OH-] = 6.10
Or,[OH-] = 10 -6.10
= 7.94 x 10^-7
pH = 7.90
Or, - log [H3O+] = 7.90
Or,[H3O+] = 10 -7.90
= 1.26 x 10^-8
4.
pH = 7.20
pOH = 14 – pH = 14 – 7.20 = 6.80
pOH = 6.80
Or, - log [OH-] = 6.80
Or,[OH-] = 10 -6.80
= 1.58 x 10^-7
pH = 7.90
Or, - log [H3O+] = 7.20
Or,[H3O+] = 10 -7.20
= 6.31 x 10^-8](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)