Acid has a pH below 7 while water has a pH of 7. A strong acid with a pH of roughly 3, HCl is. Water gets more acidic and loses pH in the range of 4-5 when HCl is added to it.
<h3>What is pH?</h3>
A substance's pH is a gauge of how basic or acidic it is. It is a measurement of the amount of H+ present in the solution. It is equivalent to the negative logarithm of the concentration of H+ ions mathematically.
A solution is acidic if its pH value is less than 7, and basic if it is greater than 7. Acids have a lower pH because they contain more H+ ions. Strong acid hydrochloric acid has a pH between 2 and 3.
Water has no charge. Water becomes acidic in pH if any acid is introduced. Therefore, if HCl is given to water, the pH of the water will change to 3-5 depending on the acid content.
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Answer:
i = 2.79
Explanation:
The excersise talks about the colligative property, freezing point depression.
Formula to calculate the freezing point of a solution is:
Freezing point of pure solvent - Freezing point of solution = m . Kf . i
Let's replace data given. (i = Van't Hoff factor, numbers of ions dissolved in solution)
48.1°C - 44°C = 0.15 m . 9.78°C/m . i
4.1°C / (0.15 m . 9.78°C/m) = i
i = 2.79
In this case, numbers of ions dissolved can decrease the freezing point of a solution, which is always lower than pure solvent.
Answer:
(NH4)2S(aq) + Pb(NO3)2(aq) --> 2NH4NO3 (aq) + PbS (s)
Answer:
B) 7.7
Explanation:
For the reaction Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq)
Kc = (CO₃²⁻) / (CrO₄²⁻)
and the Ksp given are
Ag₂CO₃ ⇒ 2 Ag⁺(aq) + CO₃²⁻(aq) Ksp₁ = (Ag⁺)²(CO₃²⁻)
Ag₂CrO₄ ⇒ 2 Ag⁺(aq)+ CrO₄²⁻(aq) Ksp₂ = (Ag⁺)²(CrO₄²⁻)
Where (...) indicate concentrations M
Notice if we divide the expressions for Ksp we get:
Ksp₁/Ksp₂ = (CO₃²⁻) / (CrO₄²⁻) = 8.5 x 10⁻¹² / 1.1 x 10⁻¹² = 7.7
which is the desired answer.
