Extreme weather e.g. snow and cold weather. Another main contributing factor is that many places have a long dry season.
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The ph after 17.0 ml of 0.15 m Koh has been added to 15 ml of 0.20 m hclo4 is <u>3.347</u>.
Titration is a commonplace laboratory technique of quantitative chemical analysis to determine the attention of an identified analyte. A reagent, termed the titrant or titrator, is ready as a trendy answer of recognized awareness and extent.
<u>Calculation:-</u>
Normality of acid Normality of base
= nMV nMV
= 1 × 0. 15 × 0.017 1 × 0. 20 ×0.015 L
= 2.55 × 10⁻³ = 3 × 10⁻³
The overall base will be high
net concentration = 3× 10⁻³ - 2.55 × 10⁻³
= 0.45 × 10⁻³
= 4.5× 10⁻⁴
pH = -log[4.5 × 10⁻⁴]
= 4 - log4.4
= <u>3.347</u>
A titration is defined as 'the manner of determining the amount of a substance A by using adding measured increments of substance B, the titrant, with which it reacts till precise chemical equivalence is completed the equivalence factor.
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1) Start by standardizing the solution of NaOH by using the solution of H2SO4 whose concentration is known.
2) Equation:
2Na OH + H2SO4 --> Na2 SO4 + 2H2O
3) molar ratios
2 mol NaOH : 1 mol H2SO4
4) Number of moles of H2SO4 in 50.0 ml of 0.0782 M solution
M = n / V => n = M*V = 0.0782 M * 0.050 l = 0.00391 mol H2SO4
5) Number of moles of NaOH
2 moles NaOH / 1mol H2SO4 * 0.00391 mol H2SO4 = 0.00782 mol NaOH
6) Concentration of the solution of NaOH
M = n / V = 0.00782 mol / 0.0184 ml = 0.425 M
7) Standardize the solution of HCl
Chemical reaction:
NaOH + HCl --> NaCl + H2O
8) Molar ratios
1 mol NaOH : 1 mol HCl
9) Number of moles of NaOH in 27.5 ml
M = n / V => n = M * V = 0.425 M * 0.0275 l = 0.01169 moles NaOH
10) Number of moles of HCl
1 mol HCl / 1mol NaOH * 0.01169 mol NaOH = 0.01169 mol HCl
11) Concentration of the solution of HCl
M = n / V = 0.01169 mol / 0.100 l = 0.1169 M
Rounded to 3 significant figures = 0.117 M
Answers:
[NaOH] = 0.425 M
[HCl] = 0.117 M
