Question

10. If 1.50 g of solid C6H-NH3Cl is dissolved in 1.0 L of water, what is the pH of the solution? Molar Mass C&H NH3CI=129.59 g/mol A. 1.93 B. 3.27 c. 7.00 D. 10.73 E. 14.00

Answer 1

here first you have to calculate the concentration of the $\small C_6H_5NH_3Cl$

molar mass of $\small C_6H_5NH_3Cl$ = 129.59 g/mol

so 1.50g of $\small C_6H_5NH_3Cl$ = $\small \frac{1.50}{129.59}$ mole = 0.0116 moles of $\small C_6H_5NH_3Cl$

so the concentration of the $\small C_6H_5NH_3Cl$ solution = = 0.0116 M

0.0116moles

along with this we need to know the value of $\small K_a$ (since the above mentioned $\small C_6H_5NH_3Cl$ salt is an acidic salt, derived from weak base $\small C_6H_5NH_2$ and strong acid $\small HCl$ , so it will show acidic character and thus we are calculating the $\small K_a$ value)

we can get that using the equation

$\small K_w =$$\small K_a$ * $\small K_b$

thus $\small K_a$ = $\small \frac{K_w}{K_b}$ ;

$\small K_b$ of $\small C_6H_5NH_2$ = 7.4 × 10⁻¹⁰ (got this value from book)

$\small K_w =$ 10^-14 (for water)

$\small K_a$ = = 1.35 * 10^-4

10-14 7.4. 10-10

rest of the answer is given in the image:

K 0 I: Cetts NH₃ C Ca Hg NH₃ + ce- Cats NH₃ + H₂O - Catts NH₂ + H₂ot 0.0116M – x Co. 0116-x) + X C: X E [H3O+] [COHEN ka= [H30+] [ Cottg NH2 x.x [Co HG NH₂ t] (0.0116-2) x is very small, so we can consider that C0.0116-2) 0.0116 ka 0.0116 putting the value of ka = 1:35x109 6 .. = 1-35x109 0.0116 ole x = 1.56810 x= 0.00125 = 1.25x10² [ Hyo+] = x = 1.25x103 pH = -log [H₃O+] = – log [H +] - i i log (1.25x10 ) - Bros - 3.25 V 202073324 09:04 LT

The answer is option B