Chapter 3: Q20P (page 62)

We can measure the concentration of HCI solution by reaction with pure sodium carbonate: $2 H^{+} + {Na}_{2} {CO}_{3} \to 2 {Na}^{+} + H_{2} 0 + 105.9884 \pm 0.0007$ required $27.35 \pm 0.04 mL$ of HCI .
(a) Find the formula mass (and its uncertainty) for ${Na}_{2} {CO}_{3}$ .
(b) Find the molarity of the HCI and its absolute uncertainty.
(c) The purity of primary standard ${Na}_{2} {CO}_{3}$ is stated to be 99.95 to 100.5wt % , which means that it can react with $(100.00 \pm 0.05) %$ of the theoretical amount of $H^{+}$ . Recalculate your answer to (b) with this additional uncertainty.

Short Answer

Expert verified

⦁ Formula mass and uncertainty of ${Na}_{2} {CO}_{3} = 0.018255 (\pm 0.093 %) mol$

⦁ Molarity and absolute uncertainty of $HCI = 0.66746 \pm 0.001$

⦁ The uncertainty would increase by 5 in the fifth decimal place. Hence, the answer would still be $0.66746 \pm 0.001$ .

Step by step solution

Given

It is given that the concentration of HCI solution can be measured by reaction with pure sodium carbonate

$2 H^{+} + {Na}_{2} {CO}_{3} \to 2 {Na}^{+} H_{2} O + {CO}_{2}$

$m ({Na}_{2} {CO}_{3}) = (0.9674 \pm 0.0009) g M ({Na}_{2} {CO}_{3}) = (105.9884 \pm 0.0007) g / mol V (HCI) = (27.35 \pm 0.04) mL$

Calculate the number of moles

(a) $H^{+}$ corresponds to HCI

Number of moles of ${Na}_{2} {CO}_{3}$ :

$n (H^{+}) = 2 n ({Na}_{2} {CO}_{3})$

Calculate the formula mass (and its uncertainty) of (Na2CO3)

$n ({Na}_{2} {CO}_{3}) = \frac{m}{M} = \frac{0.967 \pm 0.0009}{105.9884 \pm 0.0007} = \frac{0.967 \pm 0.093 %}{105.9884 \pm 0.00066 %} = 0.0091274 (\pm 0.093 %) n = 2 \times n ({Na}_{2} {CO}_{3}) = 0.018255 (\pm 0.093 %) mol$

Calculate the molarity and absolute uncetainty

(b) The molarity of HCI :

$c (HCI) = \frac{n}{V}$

$= \frac{0.018255 (\pm 0.093 %) m o l}{0.02735 \pm 0.00004 L} = 0.66740 (\pm 0.173 %) = 0.66746 (\pm 0.173 %) = 0.66746 (\pm 0.001155) = 0.667746 \pm 0.001$

Change in uncertainty

(c) In accordance with the question the uncertainty would increase by 5 in the fifth decimal place. Hence, the answer would still be $0.66746 \pm 0.001$ .