What is the pH of 0.4100 M NaCN? Be sure to use at least 4 significant figures throughout your calculations; you can only be off by 0.1 pH units.
Answer (1)
Calculate K a from the given p K a : K a = 1 0 − 9.2 = 6.30957 × 1 0 − 10 .
Calculate K b using the relationship K w = K a × K b : K b = 6.30957 × 1 0 − 10 1.0 × 1 0 − 14 = 1.58489 × 1 0 − 5 .
Determine [ O H − ] using the ICE table and the K b expression: [ O H − ] = 0.00254913 M.
Calculate the pH using pO H = − l o g [ O H − ] and p H = 14 − pO H : p H = 11.4064 .
Explanation
Problem Analysis and Hydrolysis Reaction We are asked to find the pH of a 0.4100 M NaCN solution. NaCN is the salt of a weak acid (HCN) and a strong base (NaOH). Therefore, the cyanide ion (CN-) will undergo hydrolysis in water, producing hydroxide ions (OH-) and increasing the pH of the solution. The hydrolysis reaction is: C N − ( a q ) + H 2 O ( l ) ⇌ H CN ( a q ) + O H − ( a q ) To solve this problem, we need to find the hydroxide ion concentration and then calculate the pOH and pH.
Calculating Kb First, we need to find the base dissociation constant ( K b ) for the cyanide ion. We can find this using the acid dissociation constant ( K a ) for hydrocyanic acid (HCN) and the ion product of water ( K w ). The relationship is: K w = K a × K b K b = K a K w The p K a of HCN is approximately 9.2. Therefore, we can calculate K a :
K a = 1 0 − p K a = 1 0 − 9.2 = 6.30957 × 1 0 − 10 Now we can calculate K b :
K b = 6.30957 × 1 0 − 10 1.0 × 1 0 − 14 = 1.58489 × 1 0 − 5
ICE Table and Solving for [OH-] Next, we set up an ICE (Initial, Change, Equilibrium) table to find the equilibrium concentration of O H − .
C N −
H CN
O H −
Initial
0.4100
0
0
Change
-x
+x
+x
Equilibrium
0.4100-x
x
x
Now we write the expression for K b :
K b = [ C N − ] [ H CN ] [ O H − ] = 0.4100 − x x 2 Since K b is small, we can assume that x is much smaller than 0.4100, so we can simplify the expression: 1.58489 × 1 0 − 5 = 0.4100 x 2 Solving for x: x 2 = ( 1.58489 × 1 0 − 5 ) ( 0.4100 ) = 6.49805 × 1 0 − 6 x = 6.49805 × 1 0 − 6 = 0.00254913 Therefore, [ O H − ] = 0.00254913 M
Calculating pOH and pH Now we can calculate the pOH: pO H = − l o g [ O H − ] = − l o g ( 0.00254913 ) = 2.59361 Finally, we can calculate the pH: p H = 14 − pO H = 14 − 2.59361 = 11.4064 Therefore, the pH of the 0.4100 M NaCN solution is approximately 11.4064.
Examples
Understanding pH is crucial in many real-world applications. For instance, in agriculture, knowing the pH of the soil helps farmers choose the right crops and optimize growing conditions. Similarly, in water treatment, maintaining the correct pH level ensures effective disinfection and prevents corrosion of pipes. In medicine, the pH of blood and other bodily fluids must be carefully regulated for proper physiological function. The calculations we performed are fundamental to these applications, allowing us to quantify acidity and alkalinity and make informed decisions.
