- Calculate [OH – ] and pH for
(a) 1.5 X 10 -3 M Sr(OH) 2
(b) 2.250g of LiOH in 250.0 mL of solution
(c) 1.00 mL of 0.175 M NaOH diluted to 2.00 L
(d) a soution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 X 10 -2 M Ca(OH) 2
Group of answer choices
A.) a) [OH – ]= 3.0 x 10 -3 M, pH= 11.48
b) [OH – ]= 0.3758 M, pH= 13.5750
c) [OH – ]= 8.75 x 10 -5 M, pH= 4.06
d) [OH – ]= 0.17 M, pH= 13.23
B.) a) [OH – ]= 3.0 x 10 -3 M, pH= 11.48
b) [OH – ]= 0.3758 M, pH= 13.5750
c) [OH – ]= 8.75 x 10 -5 M, pH= 9.942
d) [OH – ]= 0.17 M, pH= 13.23
C.) a) [OH – ]= 1.5 x 10 -3 M, pH= 11.18
b) [OH – ]= 0.3758 M, pH= 13.5750
c) [OH – ]= 8.75 x 10 -5 M, pH= 9.942
d) [OH – ]= 0.17 M, pH= 13.23
D.) a) [OH – ]= 3.0 x 10 -3 M, pH= 2.52
b) [OH – ]= 0.3758 M, pH= 0.425
c) [OH – ]= 8.75 x 10 -5 M, pH= 4.06
d) [OH – ]= 0.17 M, pH= 13.23
The Correct Answer and Explanation is :
To determine the hydroxide ion concentration ([OH^-]) and pH for each solution, we will analyze each case individually:
(a) 1.5 × 10⁻³ M Sr(OH)₂
Strontium hydroxide (Sr(OH)₂) is a strong base that dissociates completely in water:
[ \text{Sr(OH)}_2 \rightarrow \text{Sr}^{2+} + 2\text{OH}^- ]
Given the concentration of Sr(OH)₂ is 1.5 × 10⁻³ M, the concentration of hydroxide ions ([OH^-]) will be twice this value:
[ [OH^-] = 2 \times 1.5 \times 10^{-3} \, \text{M} = 3.0 \times 10^{-3} \, \text{M} ]
The pOH is calculated as:
[ \text{pOH} = -\log[OH^-] = -\log(3.0 \times 10^{-3}) \approx 2.52 ]
Since pH + pOH = 14, the pH is:
[ \text{pH} = 14 – 2.52 = 11.48 ]
(b) 2.250 g of LiOH in 250.0 mL of solution
First, calculate the moles of LiOH:
[ \text{Molar mass of LiOH} = 6.939 + 15.999 + 1.008 = 23.946 \, \text{g/mol} ]
[ \text{Moles of LiOH} = \frac{2.250 \, \text{g}}{23.946 \, \text{g/mol}} \approx 0.0939 \, \text{mol} ]
The volume of the solution is 250.0 mL, or 0.2500 L. Therefore, the concentration of LiOH is:
[ [\text{LiOH}] = \frac{0.0939 \, \text{mol}}{0.2500 \, \text{L}} = 0.3756 \, \text{M} ]
Since LiOH is a strong base, it dissociates completely:
[ \text{LiOH} \rightarrow \text{Li}^+ + \text{OH}^- ]
Thus, ([OH^-] = 0.3756 \, \text{M}). The pOH is:
[ \text{pOH} = -\log(0.3756) \approx 0.425 ]
And the pH is:
[ \text{pH} = 14 – 0.425 = 13.575 ]
(c) 1.00 mL of 0.175 M NaOH diluted to 2.00 L
The moles of NaOH before dilution are:
[ \text{Moles of NaOH} = 0.175 \, \text{M} \times 0.00100 \, \text{L} = 1.75 \times 10^{-4} \, \text{mol} ]
After dilution to 2.00 L, the concentration of NaOH is:
[ [\text{NaOH}] = \frac{1.75 \times 10^{-4} \, \text{mol}}{2.00 \, \text{L}} = 8.75 \times 10^{-5} \, \text{M} ]
Since NaOH is a strong base, it dissociates completely:
[ [OH^-] = 8.75 \times 10^{-5} \, \text{M} ]
The pOH is:
[ \text{pOH} = -\log(8.75 \times 10^{-5}) \approx 4.06 ]
And the pH is:
[ \text{pH} = 14 – 4.06 = 9.94 ]
(d) A solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 × 10⁻² M Ca(OH)₂
First, calculate the moles of KOH:
[ \text{Moles of KOH} = 0.105 \, \text{M} \times 0.00500 \, \text{L} = 5.25 \times 10^{-4} \, \text{mol} ]
Next, calculate the moles of Ca(OH)₂:
[ \text{Moles of Ca(OH)}_2 = 9.5 \times 10^{-2} \, \text{M} \times 0.0150 \, \text{L} = 1.425 \times 10^{-3} \, \text{mol} ]
The total volume of the solution is:
[ 5.00 \, \text{mL} + 15.0 \, \text{mL} = 20.0 \, \text{mL} = 0.0200 \, \text{L} ]
The total moles of hydroxide ions from both sources are:
[ 5.25 \times 10^{-4} \, \text{mol} + 1.425 \times 10^{-3} \,