Method for removing cationic impurities of calcium, magnesium, iron, sodium and potassium from cell grade lithium carbonate
A lithium carbonate, battery-grade technology, applied in lithium carbonate;/acid carbonate and other directions, can solve the problems of unsatisfactory impurity removal effect, unguaranteed product quality, and difficulty in controlling the content of impurity ions in lithium carbonate. To achieve the effect of reducing the loss of lithium
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Embodiment 1
[0019] The mixing ratio of water and industrial lithium carbonate is 33:1
[0020] Weigh 15.1 grams of 99% industrial lithium carbonate and 500ml of water to mix, and pass into CO at room temperature 2 When the system is clear, add 25ml of prepared 0.34mol / L trilithium ethylenediaminetetraacetic acid salt solution to the solution, stir for 30min, then heat to 85°C, and filter the precipitated lithium carbonate under reduced pressure. The obtained solid was dried at 550° C. for 3 hours, and the analysis of the impurity content of the product is shown in Table 1.
Embodiment 2
[0022] The mixing ratio of water and industrial lithium carbonate is 100:1
[0023] Weigh 5 grams of 99% industrial lithium carbonate and 500ml of water to mix, and pass into CO at room temperature 2 Until clarification, add 10ml of prepared 0.34mol / L trilithium ethylenediaminetetraacetic acid salt solution to the solution, stir for 30min, then heat to 85°C, filter the precipitated lithium carbonate under reduced pressure, and wash once with 20ml of water. The resulting solid was dried at 550 °C for 3 h. The analysis of product impurity content is shown in Table 1.
[0024] Table 1. Analysis table of product impurity content
[0025]
Embodiment 3
[0027] Preparation of Trilithium Ethylenediaminetetraacetic Acid Salt Solution
[0028] Weigh 14.6 g of EDTA and disperse it in 110 ml of water, heat to 60°C under stirring, then add 36 ml of 10% lithium hydroxide solution to obtain a clear and transparent trilithium EDTA solution.
[0029] The molar concentration of the solution is 14.6g / (292g / mol×0.146L)=0.34mol / L.
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