A rising tide is also known as an occurrence between low tide to high tide.
First, you need to find the number of moles of OH⁻ in a 250mL solution of 0.100M OH⁻. To do this, multiply 0.250L by 0.100M to get 0.025mol OH⁻. Then you use the fact that 1 mole of Sr(OH₂)·8H₂O contains 2 moles of OH⁻ which means that 0.0125mol of Sr(OH)₂·8H₂O contains 0.025mol OH⁻ (0.025/2=0.0125). Then to find the amount of Sr(OH)₂·8H₂O is needed you multiply its molar mass (265.76g/mol) by 0.0125mol to get 3.322g.
Therefore you need 3.322g of Sr(OH)₂·8H₂O.
I hope that helps. Let me know if anything is unclear.
Answer:
The pH of the solution is 4.28
Explanation:
The dissolution reaction as below
CH₃COOH ⇔ CH₃COO⁻ + H⁺
Assume the concentration of the ion, [H⁺] = a,
so [CH₃COO⁻] = a and [CH₃COOH] = 3a
Then use the formula of Ka, we get
Ka = a * a / 3a = 10^-4.76 ⇔ a = 3 x 10^-4.76 = 5.21 x 10^-5
Hence pH = -log(a) = - log(5.21 x 10^-5) = 4.28