Question

what mass (in g) of lioh would need to be dissolved in water to make 300.0 ml of a solution with a ph of 11.93?

Explanation:

Step1: Calculate pOH

$$pH + pOH=14$$
$$pOH = 14 - pH$$
$$pOH=14 - 11.93=2.07$$

Step2: Calculate \([OH^{-}]\)

$$pOH=-\log[OH^{-}]$$
$$[OH^{-}]=10^{-pOH}$$
$$[OH^{-}]=10^{- 2.07}\approx8.5\times10^{-3}\space M$$

Step3: Determine moles of \(LiOH\)

Since \(LiOH\) is a strong base and \(LiOH
ightarrow Li^{+}+OH^{-}\), the concentration of \(LiOH\) is equal to the concentration of \(OH^{-}\).
Volume \(V = 300.0\space mL=0.3000\space L\)
Moles \(n = C\times V\)
\(n_{LiOH}=[OH^{-}]\times V=(8.5\times 10^{-3}\space mol/L)\times0.3000\space L = 2.55\times10^{-3}\space mol\)

Step4: Calculate mass of \(LiOH\)

Molar mass of \(LiOH\): \(M = 23.95\space g/mol\) (\(M_{Li}=6.94\space g/mol\), \(M_{O}=16.00\space g/mol\), \(M_{H}=1.01\space g/mol\), \(M = 6.94 + 16.00+1.01=23.95\space g/mol\))
Mass \(m=n\times M\)
\(m=(2.55\times 10^{-3}\space mol)\times23.95\space g/mol\approx0.061\space g\)

Answer:

\(0.061\space g\)