Ionic product of water at 310 K is 2.7 × 10⁻¹⁴. What is the pH of neutral water at this temperature?

6.78 — For neutral water, the concentration of hydronium ions is equal to the concentration of hydroxyl ions. The ionic product of water (K w ) changes with temperature. At 310 K, K w is given as 2.7 × 10⁻¹⁴. By finding the square root of K w , we get the hydronium ion concentration, and taking the negative logarithm gives the pH. Alternatively, one can calculate pK w and divide it by 2 to find the pH of neutral water.

Ionic product of water at 310 K is 2.7 × 10⁻¹⁴. What is the pH of…

Question 62 of 69beginner🔧 ApplyNumerical1 mark

Ionic product of water at 310 K is $2.7 \times 10^{-14}$ . What is the pH of neutral water at this temperature?

Correct Answer

6.78

Exercise: EXERCISES | Q: 6.65 | (Page 46)

For More Understanding

Explanation

For neutral water, the concentration of hydronium ions is equal to the concentration of hydroxyl ions. The ionic product of water ( $K_w$ ) changes with temperature. At 310 K, $K_w$ is given as $2.7 \times 10^{-14}$ . By finding the square root of $K_w$ , we get the hydronium ion concentration, and taking the negative logarithm gives the pH. Alternatively, one can calculate $pK_w$ and divide it by 2 to find the pH of neutral water.

Solution Steps

Step 1: For neutral water, $[H_3O^+] = [OH^-]$ . The ionic product of water is given by $K_w = [H_3O^+] [OH^-] = [H_3O^+]^2$ .
Step 2: Given $K_w = 2.7 \times 10^{-14}$ at 310 K. We can find $pK_w = -\log K_w = -\log(2.7 \times 10^{-14}) = 14 - \log(2.7) = 14 - 0.43 = 13.57$ .
Step 3: For a neutral solution, $\text{pH} = \text{pOH}$ . Since $\text{pH} + \text{pOH} = pK_w$ , we have $2 \times \text{pH} = pK_w$ .
Step 4: Therefore, $\text{pH} = \frac{pK_w}{2} = \frac{13.57}{2} = 6.785 \approx 6.78$ .

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