Question

Thiamine hydrochloride (C12H18ON4SCl2) is a water-soluble form of thiamine (vitamin B1; Ka = 3.37×10−7). How many grams of the hydrochloride must be dissolved in 10.00 mL of water to give a pH of 3.50?

Answer 1

Answer:

$m_{HA}=0.784g$

Explanation:

Hello,

In this case, considering the dissociation of thiamine hydrochloride:

$C_{12}H_{18}ON_4SCl_2 \rightleftharpoons H^++C_{12}H_{17}ON_4SCl_2^-$

It is convenient to write it as:$C_{12}H_{18}ON_4SCl_2+H_2O \rightleftharpoons H_3O^++C_{12}H_{17}ON_4SCl_2^-$

Or:

$HA+H_2O\rightleftharpoons H_3O^++A^-$

With which the Henderson-Hasselbalch equation is applied:

$pH=pKa+log(\frac{[A^-]}{[HA]} )$

Therefore:

$log(\frac{[A^-]}{[HA]} )=3.50-[-log(3.37x 10^{-7})]=3.50-6.47=-2.97}\\\\\frac{[A^-]}{[HA]} =10^{-2.97}=1.07x10^{-3}$

$[A^-]=1.07x10^{-3}}[HA]$

Thus, as the pH equals the concentration of hydrogen which also equals the concentration of the conjugate base, one obtains:

$[H]^+=[A^-]=10^{-pH}=10^{-3.50}=3.16x10^{-4}M$

Now, solving for the concentration of acid ([HA]):

$[HA]=\frac{3.16x10^{-4}M}{1.07x10^{-3}} =0.296M$

Finally, the mass turns out:

$m_{HA}=0.01000L*0.296\frac{mol}{L}*\frac{265.35g}{1mol}\\ m_{HA}=0.784g$

Best regards.