Titration help

PODHEC1

13

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Is the answer to b 7.87 gdm-3

(edited 6 years ago)

Reply 1

username2889812

19

Original post by HN786

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Is the answer to b 7.87 gdm-3

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Reply 2

PODHEC1

OP

13

Original post by Iluv2edgedsword

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This work?

Reply 3

Dysf(x)al

20

Original post by HN786

This work?

I got 3.94 gdm^-3 (half of what you got):

0.0528 moldm^-3 * 0.025 dm^3 = 0.001321 mol of sodium carbonate
Each mole of sodium carbonate requires 2 moles of hydrochloric acid, so 0.001321*2 = 0.2642 moles of hydrocholoric acid were used.
0.2642 mol / 0.0245 dm^3 = 0.1078 moldm^-3 hydrochloric acid
RFM of hydrochloric acid is 1 + 35.5 = 36.5 (your periodic table may have slightly different values - use those).
0.1078 moldm^-3 * 36.5 gmol^-1 = 3.94 gdm^-3.

Reply 4

PODHEC1

OP

13

Original post by TheMindGarage

I got 3.94 gdm^-3 (half of what you got):

0.0528 moldm^-3 * 0.025 dm^3 = 0.001321 mol of sodium carbonate
Each mole of sodium carbonate requires 2 moles of hydrochloric acid, so 0.001321*2 = 0.2642 moles of hydrocholoric acid were used.
0.2642 mol / 0.0245 dm^3 = 0.1078 moldm^-3 hydrochloric acid
RFM of hydrochloric acid is 1 + 35.5 = 36.5 (your periodic table may have slightly different values - use those).
0.1078 moldm^-3 * 36.5 gmol^-1 = 3.94 gdm^-3.

Why would you multiply the moles of sodium carbonate by 0.025

Reply 5

Luxtra

2

Hi, just to jump in. They multiplied the concentration (0.0528M) by the volume (0.025dm3 or 25cm3) in order to get the moles, as moles = conc * vol

Reply 6

Maker

20

1.4g = 1.4/106 = 0.013 moles

0.013 moles in 250ml

therefore there is 4x0.013 moles of of sodium carbonate in 1dm^3 = 0.052 mol/dm^3

25ml of 0.052 molar solution of sodium carbonate reacts with 24.5 ml HCl.

The molarity of the acid is 25x2x0.052/24.5 = 0.106mol/dm^3

There is 0.106x36.5g = 3.87g/dm^3

My answer has fewer rounding errors.

(edited 6 years ago)

Reply 7

Dysf(x)al

20

Original post by TheMindGarage

I got 3.94 gdm^-3 (half of what you got):

0.0528 moldm^-3 * 0.025 dm^3 = 0.001321 mol of sodium carbonate
Each mole of sodium carbonate requires 2 moles of hydrochloric acid, so 0.001321*2 = 0.2642 moles of hydrocholoric acid were used.
0.2642 mol / 0.0245 dm^3 = 0.1078 moldm^-3 hydrochloric acid
RFM of hydrochloric acid is 1 + 35.5 = 36.5 (your periodic table may have slightly different values - use those).
0.1078 moldm^-3 * 36.5 gmol^-1 = 3.94 gdm^-3.

Because you used 25cm^3 of it. 25cm^3 = 0.025dm^3.