Question

a chemist dissolves 440. mg of pure potassium hydroxide in enough water to make up 150. ml of solution. calculate the ph of the solution. (the temperature of the solution is 25°c.)
round your answer to 3 significant decimal places.

Explanation:

Step1: Calculate the molarity of \(KOH\) solution

• First, convert the mass of \(KOH\) to moles. The molar mass of \(KOH\) is \(M=(39 + 16+1)\space g/mol = 56\space g/mol\). The mass of \(KOH\) is \(m = 440\space mg=0.440\space g\). Using the formula \(n=\frac{m}{M}\), we have \(n=\frac{0.440\space g}{56\space g/mol}\approx7.86\times 10^{-3}\space mol\).
• Then, convert the volume of the solution to liters. \(V = 150\space mL = 0.150\space L\).
• Use the formula for molarity \(c=\frac{n}{V}\). So, \(c=\frac{7.86\times 10^{-3}\space mol}{0.150\space L}\approx0.0524\space mol/L\).

Step2: Determine the \(OH^-\) concentration

Since \(KOH\) is a strong base and dissociates completely as \(KOH=K^{+}+OH^{-}\), the concentration of \(OH^-\) is equal to the concentration of \(KOH\). So, \([OH^{-}]=0.0524\space mol/L\).

Step3: Calculate the \(pOH\)

Use the formula \(pOH =-\log[OH^{-}]\). Then \(pOH=-\log(0.0524)\approx1.281\).

Step4: Calculate the \(pH\)

At \(25^{\circ}C\), \(pH + pOH=14\). So, \(pH=14 - pOH\). Substitute \(pOH = 1.281\) into the formula, we get \(pH=14 - 1.281=12.719\).

Answer:

\(12.7\)