Calculate the pH of a 0.35 M solution of C2H$NH3CI (Kb for C2HSNH2-5.6 x 104). R

Question

Calculate the pH of a 0.35 M solution of C2H$NH3CI (Kb for C2HSNH2-5.6 x 104). Record your pH value to 2 decimal places. 0.54 For salt solutions, remember: A. +1 and +2 metal ions generally have no acidic/basic properties. B. the conjugate bases of weak acids are weak bases (1 Kb> 10-14. C, the conjugate bases of strong acids are worse bases than water (Kb acids are worse bases than water (Kp 1014) 1 D. the conjugate acids of weak bases are weak acids (1 > K > 1014) Once you have identified the acid/base properties of the ions, then dolve the appropriate problem to determine the pH. If you have an ion which behaves as a weak base, write out the Ko reaction for the ion and solve the equilibium problem in order to determine pH. If you have an ion which behaves as a weak acid, write out the Ka reaction for the ion and solve the equilibium problem in order to determine pH. If you have an ionic compound whose ions neither have acidic or basic properties, then the pH will be that of neutral water (pH=7.00). Submit Answer Incorrect. Tries 4/45 Previnus Triac

Explanation / Answer

Here is the chemical reaction of the hydrolysis of C2H5NH3+:

C2H5NH3+ + H2O = C2H5NH2 + H3O+

Here is the Ka expression for C2H5NH3+:

We can then substitute values into the Kb as:

Ignoring the - x:

x = 1.96 x 10¯4 = [H3O+]

so pH would be:

pH = - log 1.96 x 10¯4 = 3.71

[C2H5NH2] [H3O+] Kb = ---------------- [C2H5NH3+]

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