Print Print HCalculator Perlodlo Table Question 6 of 16 Incorrect Incorrect Sapl

Question

Print Print HCalculator Perlodlo Table Question 6 of 16 Incorrect Incorrect Sapling Learning Map Determine the pH of a solution when 26.0 mL ofo.027 M HNo, is mixed with 19.5 mL of A) 0.0070 M NaOH. Number pH 1.91 B) distilled water. Number pH 0.397 C00.0100 MHCI. Number pH 1.89 D) 0.084 M KOH. Mixing the KoH solution with the original HNO solution results in an Number excess number of moles of oH which are found by subtracting the moles of H in the onginal solution from the moles of oH in the pH 12.39 added solution when calculating the OH concentration of the AG Previous Try Again Next Ent tL about us careers partners privacy policy terms of use

Explanation / Answer

millimoles of HNO3 = 26 x 0.027 = 0.702

A) millimoles of NaOH added = 19.5 x 0.007 = 0.1365

0.702 - 0.1365 = 0.5655

total volume = 26 + 19.5 = 45.5

[HNO3] = 0.5655 / 45.5 = 0.0124 M

pH = - log [H+]

pH = - log [0.0124]

pH = 1.91

part B) total volume = 45.5 mL

[HNO3] = 0.702 / 45.5 = 0.0154 M

pH = - log [H+]

pH = - log [0.0154]

pH = 1.82

part C) millimoles of HCl added = 0.01 x 19.5 = 0.195

total millimoles of acid = 0.195 + 0.702 = 0.897

[H+] = 0.897 / 45.5 = 0.0197 M

pH = - log [H+]

pH = - log [0.0197]

pH = 1.70

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