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Sebanyak 100 mL larutan Ca ( NO 3 ​ ) 2 ​ 0,05 M dicampurkan dengan 100 mL larutan Na 2 ​ CO 3 ​ 0,05 M dengan reaksi: Ca ( NO 3 ​ ) 2 ​ ( a q ) + Na 2 ​ CO 3 ​ ( a q ) → CaCO 3 ​ ( s ) + 2 NaNO 3 ​ ( a q ) Jika K sp ​ CaCO 3 ​ = 9 × 1 0 − 9 , massa yang mengendap sebanyak .... ( 0.9 ​ = 0 , 95 )

Sebanyak 100 mL larutan  0,05 M dicampurkan dengan 100 mL larutan  0,05 M dengan reaksi:



Jika , massa yang mengendap sebanyak .... ()

  1. 0,025 gramspace

  2. 0,100 gramundefined 

  3. 0,25 gramundefined 

  4. 0,50 gramundefined 

  5. 1,000 gramundefined 

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Y. Rochmawatie

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah D.

jawaban yang benar adalah D.

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Pembahasan

Pembahasan
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Tahap 1. Menentukan apakah akan mengendap atau tidak. Diketahui : setelah pencampuran setelah pencampuran akan mengendap karena . Tahap 2. Mencari mol masing-masing senyawa. Tahap 3. Mencari massa yang mengendap. Jadi, jawaban yang benar adalah D.

Tahap 1. Menentukan apakah begin mathsize 14px style Ca C O subscript 3 end style akan mengendap atau tidak.

Diketahui :

begin mathsize 14px style space left square bracket Ca open parentheses NO subscript 3 close parentheses subscript 2 right square bracket equals straight M subscript 1 equals 0 comma 05 space M V space Ca left parenthesis NO subscript 3 right parenthesis subscript 2 equals straight V subscript 1 equals 100 space mL space space space left square bracket Na subscript 2 CO subscript 3 right square bracket equals straight M subscript 1 equals 0 comma 05 space straight M space space space straight V space Na subscript 2 CO subscript 3 equals straight V subscript 1 equals 100 space mL  straight V subscript 2 equals V space Ca left parenthesis NO subscript 3 right parenthesis subscript 2 plus straight V space Na subscript 2 CO subscript 3 space space space space equals 100 space mL plus 100 space mL space space space space equals 200 space mL end style  

begin mathsize 14px style open square brackets Ca open parentheses N O subscript 3 close parentheses subscript 2 close square brackets end style setelah pencampuran

begin mathsize 14px style space space space space space space space space space space space space space space space space space space straight M subscript 1 straight V subscript 1 equals straight M subscript 2 straight V subscript 2 0 comma 05 space straight M cross times 100 space mL equals straight M subscript 2 cross times 200 space mL space space space space space space space space space space space space space space space space space space space space space space straight M subscript 2 equals 0 comma 025 space straight M end style  

begin mathsize 14px style Ca open parentheses N O subscript 3 close parentheses subscript 2 rightwards harpoon over leftwards harpoon space space Ca to the power of 2 plus space space end exponent plus 2 N O subscript 3 to the power of minus sign 0 comma 025 space M space space space space space space 0 comma 025 space M space space space 0 comma 05 M end style 

begin mathsize 14px style open square brackets Na subscript 2 C O subscript 3 close square brackets end style setelah pencampuran 

undefined  

begin mathsize 14px style Na subscript 2 C O subscript 3 yields space 2 Na to the power of plus sign space space plus C O subscript 3 to the power of 2 minus sign end exponent 0 comma 025 space M space space space space 0 comma 05 space M space space space space 0 comma 025 space M end style 

begin mathsize 14px style straight Q subscript sp space CaCO subscript 3 equals open square brackets Ca to the power of 2 plus end exponent close square brackets open square brackets CO subscript 3 to the power of 2 minus end exponent close square brackets space space space space space space space space space space space space space space space space space space equals 0 comma 025 space straight M cross times 0 comma 025 space straight M space space space space space space space space space space space space space space space space space space equals 6 comma 25 cross times 10 to the power of negative 4 end exponent end style 

begin mathsize 14px style straight K subscript sp space CaCO subscript 3 equals 9 cross times 10 to the power of negative 9 end exponent end style

undefined akan mengendap karena begin mathsize 14px style Q subscript sp greater than K subscript sp end style.

Tahap 2. Mencari mol masing-masing senyawa.

begin mathsize 14px style mol space Ca left parenthesis NO subscript 3 right parenthesis subscript 2 equals 0 comma 1 space cross times space 0 comma 05 equals 5 space cross times space 10 to the power of negative 3 space end exponent mol space space space mol space Na subscript 2 CO subscript 3 equals space 0 comma 1 space cross times space 0 comma 05 equals 5 space cross times space 10 to the power of negative 3 space end exponent mol end style   

 

Tahap 3. Mencari massa undefined yang mengendap. 

begin mathsize 14px style straight M subscript straight r space CaCO subscript 3 equals left parenthesis straight A subscript straight r space Ca right parenthesis plus left parenthesis straight A subscript straight r space straight C right parenthesis plus left parenthesis 3 cross times straight A subscript straight r space straight O right parenthesis straight M subscript straight r space CaCO subscript 3 equals 40 plus 12 plus left parenthesis 3 cross times 16 right parenthesis straight M subscript straight r space CaCO subscript 3 equals 52 plus 48 straight M subscript straight r space CaCO subscript 3 equals 100 space straight g space mol to the power of negative 1 end exponent  massa space CaCO subscript 3 equals straight n space CaCO subscript 3 space cross times space straight M subscript straight r space CaCO subscript 3 massa space CaCO subscript 3 equals 5 space cross times space 10 to the power of negative 3 end exponent space mol space cross times space 100 space straight g space mol to the power of negative 1 end exponent massa space CaCO subscript 3 equals 0 comma 5 space gram end style   

Jadi, jawaban yang benar adalah D.

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